elements of dynamics s l loney

A TREATISE

ON

ELEMENTAEY DYNAMICS

C. J. CLAY AND SONS,

CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,

AVE MARIA LANE.

CAMBRIDGE: DEIGHTON, BELL, AND CO.

LEIPZIG: F. A. BROCKHAU8.

\\

A TKEATISE

ON

ELEMENTAKY DYNAMICS^

BY

\ '

S. L. LONEY, M.A.

FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE.

CAMBRIDGE:AT THE UNIVERSITY PRESS.

1889

[All Rights reserved.]

PBINTBD BY C. J. CLAY, M.A. AND SONS,

AT THE UNIVERSITY PRESS.

PEEFACE.

IN the following work I have attempted to write a

fairly complete text-book on those parts of Dynamicswhich can be treated without the use of the Infinitesimal

Calculus..

The book is intended for the use of beginners, but it is

probable that most students would find it advisable, at

any rate on the first reading of the subject, to confine

their attention to the examples given in the text, reserving

most of the questions at the ends of the chapters for a

second reading. Most students would also find it ad-

visa,ble to omit Chaj^-III., the last four articles of Chap.

IV., and Chap. IX., until the rest of the book has been

fairly mastered.

I have ventured to alter slightly the time-honoured

enunciation of Newton's Second Law of Motion.

I have used the property of the hodograph in the

chapter on normal acceleration and in dealing with the

motion of a particle in a conic section with an acceleration

directed towards the focus.

L. D. I)

VI PREFACE.

In treating cycloidal and pendulum motion I have first

introduced the student to the conception of simple har-

monic motion and thus have avoided much of the dreary

work usually to be found in this portion of the subject.

I must express my gratitude to Mr C. SMITH, M.A.,

Fellow and Tutor of Sidney Sussex College, Cambridge,

for much kind criticism and encouragement; also to Mr

H. C. ROBSON, M.A., of Sidney Sussex College, and other

friends.

I have spared no pains to ensure the accuracy of the

answers to the examples, and hope that not many serious

errors will be found. I could not, however, burden myfriends with the large amount of drudgery involved in

their verification and so was obliged to rely on my own

unaided resources.

For any corrections, or suggestions for improvement,

I shall be very grateful.

S. L. LONEY.

LONSDALE EOAD, BARNES. S. W.

July 17, 1889.

CONTENTS.

CHAPTER I.

UNIFORM AND UNIFORMLY ACCELERATED MOTION.PAGE

Dynamics. Definition and Subdivisions 1

Speed 2

Velocity 4

Parallelogram of Velocities 7

Component and resultant velocities 8

Triangle and Parallelepiped of Velocities 10

Examples 12

Change of velocity and acceleration . . . . . .15Parallelogram of Accelerations 16

Relative Velocity. Examples . 18

Angular and Areal Velocity. Examples 24

Formulae to determine the motion when the acceleration and initial

velocity are given. Examples 28

Acceleration of falling bodies. Morin's experiment ... 33

Vertical motion under gravity. Examples 34

Motion on a smooth inclined plane. Examples .... 38

Time down chords of a vertical circle 40

Lines of quickest descent 41

Graphic Methods. Velocity-Time Curve. Acceleration-Time

Curve 46

Examples on Chapter I. . 50

viii CONTENTS.

CHAPTER II.

THE LAWS OF MOTION.PAGE

Definitions 5&

Enunciation of the Laws of Motion 60

The relation P=mf .62Definition of the unit of force 63

Absolute and gravitation units .64The weight of a body is proportional to its mass .... 64

Distinction between mass and weight 65

Weighing by scales and a spring-balance 66

Examples 67

Physical independence of forces . . 69

Parallelogram of Forces - .70Examples of the Third Law of Motion 71

Motion of two particles connected by a string placed over a smooth

pulley72

Atwood's machine . . 73

Examples 75

Impulse and impulsive forces . 81

Impact of two bodies 84

Motion of a shot and gun 85

Work and Power 86

Examples 89

Kinetic and Potential Energy 92

Conservation of Energy 93

Graphic Methods. Force-Space Curve 95

Motion of the centre of inertia of a system of bodies ... 96

Conservation of Momentum . 100

Miscellaneous Examples 101

Examples on Chapter II. 103

CHAPTER III.

THE LAWS OF MOTION (continued). MISCELLANEOUSEXAMPLES.

Motion of two pulleys, each carrying masses, connected by a string

passing over a fixed pulley 114

Motion of a wheel-and-axle . . . . . . . . 116

CONTENTS. IX

PAGE

Motion of a movable wedge with masses upon its faces . . .117Ratio of velocity of a cannon-ball when the cannon is free to move

to the velocity when the cannon is fixed 119

Perforation of a plate by a shot 120

Pile-driving 121

Man leaping from a movable platform 122

Motion of a heavy elastic ring upon a smooth cone . . . 123

Starting of a goods train with the couplings slack .... 123

Motion of heavy strings 125

Transmission of power by belts and shafting 128

Examples on Chapter III 129

CHAPTER IV.

UNITS AND DIMENSIONS.

Fundamental and derived units 143

The unit of velocity varies directly as the unit of length, and

inversely as the unit of time 145

The unit of acceleration varies directly as the unit of length, and

inversely as the square of the unit of time .... 146

Definition of dimensions 148

Dimensions of Velocity, Force, Work, etc 148

Examples 151

Verification of formulae by means of counting the dimensions . 156

Attraction units 158

Astronomical unit of mass 161

Table of dimensions and values of fundamental quantities . . 162

Examples on Chapter IV . 164

CHAPTER Y.

PROJECTILES.

Definitions and properties of a parabola 167The path of a projectile is a parabola 170The velocity at any point is that due to a fall from the directrix . 171

Greatest height and range on a horizontal plane. Maximum range 173

X CONTENTS.

PAGE

Two directions of projection for a given range .... 174

Latus-rectum of the path 175

Velocity and direction of motion at any time . . . . .175Focus and directrix of path . 176

Equation to the path . . . .177Examples 177

Eange on an inclined plane. Maximum range .... 181

Motion upon an inclined plane 183

Examples 184

Geometrical construction for the path 185

Envelope of paths . 186

Geometrical construction for maximum range .... 188


Examples on Chapter V. 191

CHAPTER VI.

COLLISION OF ELASTIC BODIES.

Elasticity '202

Newton's Experimental Law 203

Impact on a smooth fixed plane 205

Direct impact of two spheres 207

Oblique impact of two spheres ....... 209

Examples 212

Loss of Kinetic Energy by impact . . . . . . . 217

Action between two elastic bodies during their collision . . , 219

Impact of a particle on a rough plane 220


Examples on Chapter VI 227

CHAPTER VII.

THE HODOGRAPH AND NORMAL ACCELERATIONS.

Definition of the hodograph 241

The velocity in the hodograph represents the acceleration in the

path . .242

CONTENTS. XI

PAGE

Normal acceleration of a point moving in a circle with uniform

speed 243

Normal and tangential accelerations in any curve .... 245' '

Centrifugal Force" 247

Examples 248

The conical pendulum 250

Railway carriage on a curved line 252

Rotating sphere 254

Revolving string . 254

Examples 255

On a smooth curve, under gravity only, the change of velocity is

that due to the vertical distance through which the particle

moves 257

Newton's Experimental Law 260

Motion on the outside of a vertical circle ..... 261

Motion in a vertical circle 262

Effect of the rotation of the earth upon the apparent weight of a

pariicle 265

Examples on Chapter VII 268

CHAPTER VIII.

SIMPLE HARMONIC MOTION. CYCLOIDAL ANDPENDULUM MOTIONS.

Definition and investigation of simple harmonic motion . . . 277

Extension to motion in a curve . . . ..

. . . 281

Examples 282

Definition and properties of a cycloid . . . . . . 284

Isochronism of a cycloid 287

Simple pendulum 288

Small oscillations in a vertical circle 289

Seconds pendulum 290

Simple equivalent pendulum 291

Effect on the time of oscillation of small changes in the value of

"g" and of small changes in the length of the pendulum . 292

Finding"g"by means of the pendulum 294

Verification of law of gravity by means of the moon's motion . 295

Examples on Chapter VIII 295

Xll CONTENTS.

CHAPTER IX.

MOTION OF A PAETICLE ABOUT A FIXEDCENTBE OF FOECE.

PAGE

Definition of the moment of a velocity 300

The moment of the velocity of a particle about a fixed centre, to

which its acceleration is always directed, is constant . . 301

Motion in an ellipse about the centre 302

Motion in a conic section about the focus 305

Kepler's Laws 308

Effect of disturbing forces on the path of a particle . . . 309

Effect of a change in the absolute value of the acceleration . . 310

Motion in a straight line with an acceleration varying inversely as

the square of the distance 310

Motion of a projectile, variations of gravity being considered . . 311


Examples on Chapter IX. . . 315

ANSWERS TO THE EXAMPLES , 321

CHAPTER I.

UNIFORM AND UNIFORMLY ACCELERATED MOTION.

1. Dynamics is the science which treats of the

action of force on bodies.

When a body is acted on by one or more forces, their

effect is either (1) to compel rest or prevent motion, or (2)

to produce or change motion.

Dynamics is therefore conveniently divided into two

portions, Statics and Kinetics.

In Statics the subject of the equilibrium, or balancing,of forces is considered

;in Kinetics is discussed the action

of forces in producing, or changing, motion.

The present book treats only of Kinetics.

The branch of Pure Mathematics which deals with the

geometry of the motion of bodies, without any reference

to the forces which produce or change the motion, is

called Kinematics. It is a necessary preliminary to

Dynamics.The present chapter Avill be devoted to Kinematics

;

in the next chapter we shall enter on Dynamics.

2. The position of a point in a plane is generally

determined in one of two ways. We may choose a point

L. D. 1

2 UNIFORM AND UNIFORMLY

fixed in the plane, and a fixed straight line Ox through

;then the position of a point P is known, when the

distance OP and the angle xOP are known.

Or again we may choose a fixed point and two fixed

straight lines Ox, Oy drawn from;from a point P draw

PM, PN parallel to Oy, Ox respectively to meet Ox, Oy in

M, N. The position of P is now known if the distances

MP, NP are known.

Here we observe that to fix the position of a point in

a plane we must have two quantities given, viz. either

a length and a direction, or two lengths in two fixed

directions.

3. Change of position. If at any instant the position

of a moving point is P, and at any subsequent time it is Q,

then PQ is the change of its position in that time.

A point is said to be in motion when it changes its

position.

The path of a moving point is the curve drawn throughall the successive positions of the point.

4. Speed. Def. The speed of a moving point is the

rate at which it describes its path.

A point is said to be moving with uniform speed when

it moves through equal lengths in equal times, however

small these times may be.

Suppose a train describes 30 miles in each of several consecutive

hours. We are not justified in saying that its speed is uniform unless

we know that it describes half a mile in each minute, 44 feet in each

second, one-millionth of 30 miles in each one-millionth of an hour, and

so on.

When uniform, the speed of a point is measured bythe distance passed over by it in a unit of time

;when

ACCELERATED MOTION. 3

variable, by the distance which would be passed over bythe point in a unit of time, if it continued to move duringthat unit of time with the speed it has at the instant

under consideration.

By saying that a train is moving with a speed of 40

miles an hour, we do not mean that it has gone 40 miles

in the last hour, or that it will go 40 miles in the next

hour;but that if its speed remained constant for one

hour then it would describe 40 miles in that hour.

5. The definition of the measure of the speed of a point at anyinstant may also be stated as follows :

Let a be the length of the portion of the path passed over by the

moving point in the time r following the instant under consideration ;

then the limiting value of the ratio -,as the time r is taken indefinitely

small, is the measure of the speed at that instant.

6. The units of length and time usually employed in

England are a foot and a second.

In scientific measurements the unit of length usually

employed is the centimetre, which is the one hundredth

part of a metre.

One metre = 39'37 inches approximately. A deci-

metre isj-'jyth,

and a millimetre YFjyu^h ^ a metre.

7. The unit of speed is the speed of a point which

moves over a unit of length in a unit of time. Hence the

unit of speed depends on these two units, and if either, or

both of them, be altered the unit of speed will also, in

general, be altered.

8. If a point is moving with speed u, then in each

unit of time the point moves over u units of length.

.'.in t units of time the point passes over ut units of

length.

12

4 VELOCITY.

Hence the distance s passed over by a point which

moves with speed u for time t is given by s = ut.

Ex. 1. Find the speed of the centre of the earth in metres per

second, assuming that it describes a circle of radius 93,000,000 miles

in 365 days.

Ex. 2. Find the speed of light in miles per second, if it takes 8

minutes to describe the distance from the sun to the earth.

Ex. 3. Find the speed, in metres per second, arising from the rotation

of the earth, of a point in latitude 60.

9. Displacement. The displacement of a moving

point is its change of position. To know the displacement

of a moving point we must know both the length of the

line joining the two positions of the point, and also the

direction. Hence the displacement of a point involves both

magnitude and direction.

Ex. 1. A man walks 3 miles due east, and then 4 miles due north ;

find his displacement.

Ans. 5 miles at an angle tan- 1 north of east.

Ex. 2. A ship sails 1 mile due south and then *J2 miles south-west ;

shew that its displacement is ^5 miles in a direction tan-1 \ west ot

south.

10. Velocity. Def. The velocity of a moving point

is the rate of its displacement.

A velocity therefore possesses both magnitude and

direction.

A point is said to be moving with uniform velocity,

when it is moving in a constant direction, and passes over

equal lengths in equal times, however small these times

may be.

When uniform the velocity of a moving point is

measured by its displacement per unit of time; when

VELOCITY. 5

variable, it is measured, at any instant, by the displace-

ment that the moving point would have in a unit of time,

if it moved during that unit of time with the velocity

which it has at the instant under consideration.

11. The definition of the measure of the velocity of a point at a

given instant may also be stated as follows :

Let d be the displacement of a moving point in the time r following

the instant under consideration;then the limiting value of -

,as the

time T is taken indefinitely small, is the measure of the velocity at that

instant.

12. It will be noted that when the moving point is

moving in a straight line, the velocity is the same as the

speed. The word velocity is often used to express what

we in Art. 4 call the speed of the moving point. It seems

however desirable, at any rate for the beginner, to care-

fully distinguish between the two.

If the motion be not in a straight line the velocity is

not the same as the speed. For example, suppose a pointto be describing a circle uniformly so that it passes over

equal lengths of the arc in equal times however small;

its direction of motion (viz. the tangent to the circle) is

different at different points of the circumference;hence

the velocity of the point (strictly so called) is variable,

whilst its speed is constant.

13. The unit of velocity is the velocity of a point

which undergoes a displacement equal to a unit of lengthin a unit of time.

When we say that a moving point has velocity v, wemean that it possesses v units of velocity, i.e. that it would

undergo a displacement, equal to v units of length, in the

unit of time.

6 VELOCITY.

If the velocity of a moving point in one direction be

denoted by v, an equal velocity in an opposite direction is

necessarily denoted by v.

14. Since the velocity of a point is known when its

direction and magnitude are both known, we can con-

veniently represent the velocity of a moving point by a

straight line AB;thus when we say that the velocities of

two moving points are represented in magnitude and

direction by the straight lines AB, CD, we mean that

they move in directions parallel to the lines drawn from

A to B, and C to D respectively, and with velocities which

are proportional to the lengths A B and CD.

15. A body may have simultaneously velocities in

two or more different directions. One of the simplest

examples of this is when a person walks on the deck of a

ship in motion from one point of the deck to another. Hehas a motion with the ship, and one along the surface of

the ship, and his motion in space is clearly different from

what it would have been had either the ship remained at

rest, or had the man stayed at his original position on the

deck.

Again consider the case of a ship steaming with its bow

pointing in a constant direction, say due north, whilst a

current carries it in a different direction, say south-east,

and suppose a sailor is climbing a vertical mast of the

ship. The actual change of position and the velocity of

the sailor clearly depend on three quantities, viz. the rate

and direction of the ship's sailing, the rate and direction of

the current, and the rate at which he climbs the mast.

His actual velocity is said to be "compounded" of these

three velocities.

PARALLELOGRAM OF VELOCITIES. 7

In the following article we shew how to find the

velocity which is equivalent to two velocities given in

magnitude and direction.

16. Theorem. Parallelogram of Velocities. //a moving point possess simultaneously velocities which are

represented in magnitude and direction by the two sides

of a parallelogram drawn from a point, they are equiva-

lent to a velocity which is represented in magnitude and

direction by the diagonal of the parallelogram passing

through the point.

Let the two simultaneous velo-

cities be represented by the lines

A B, A C and let their magnitudesbe n, v.

Complete the parallelogram

BACD.Then we may imagine the motion of the point to be

along the line AB with the velocity u, whilst the page of

this book moves parallel to AC with velocity v. In the

unit of time the moving point will have moved through a

distance AB along the line AB, and the line AB will have

in the same time moved into the position CD so that the

moving point will now be at D.

Now, since the two coexistent velocities are constant

in magnitude and direction, the velocity of the point from

A to D must be also constant in magnitude and direction;

hence AD is the path described by the moving point in

the unit of time.

Hence AD represents in magnitude and direction the

velocity which is equivalent to the velocities represented

by AB and AC.

8 RESOLUTION AND

17. Def. The velocity which is equivalent to two or

more velocities is called their resultant, and these velocities

are called the components of this resultant.

The resultant of two velocities u, v in directions which

are inclined to one another at a given angle a may be easily

obtained.

For in Fig. Art. 16, let AB, AC represent the velocities

u, v, so that L BAC= a.

Then AD'1 = AB* + 3D2 - 2AB . BD cos ABD.Hence if we represent the resultant velocity AD by w,

we have' wz

u* + v*+ 2uv cos a, since / ABD = TT y.

sin BAD BDAlso

sin A&& AB '

sin 6 v ,=,where L BAD = 6

;

sin (a 0) u

v sin a/. tan 6 =

u + v cos -JL

Hence the resultant of two velocities u, v inclined to one

another at an angle v., is a velocity Jtf + v" 4- 2uv cos a

v *?? 7? fy.

inclined at an anqle tan'1

to the direction of theu + v cos a

velocity u.

The direction of the resultant velocity may also be obtained as follows;

draw DE perpendicular to AB to meet it, produced if necessary, in E ;

we then have

DE BD sin EBD v sin atan DAB= ="

AE AB +BDcosEBD u + vcosa'

18. A velocity can be resolved into two componentvelocities in an infinite number of ways. For an infinite

number of parallelograms can be described having a line

COMPOSITION OF VELOCITIES. 9

AD as diagonal ;and if ACDB be any one of these the

velocity AD is equivalent to the two component velocities

AB and AC.

The most important case is when a velocity is to be

resolved into two velocities in two directions at right

angles, one of these directions being given. When we

speak of the component of a velocity in a given direction it

is to be understood that the other direction in which the

given velocity is to be resolved is perpendicular to this

given direction.

Thus suppose we wish to resolve a velocity u repre-

sented by AD into two componentsat right angles to one another, one

of these components being along a

line AB making an angle 6 with

AD.Draw DB perpendicular to AB,

and complete the rectangle ABDC.Then the velocity AD is equivalent to the two com-

ponent velocities AB, AC.

Also AB = AD cos 6 = u cos 0,

and A C = AD sin0 = u sin 9.

We thus have the following important

Theorem. A velocity u is equivalent to a velocity

u cos along a direction making an angle 6 with its own

direction together with a velocity u sin 6 perpendicular to

the direction of the first component.

Ex. 1. A man is walking in a north-easterly direction with a velocity

of 4 miles per hour ;find the components of his velocity in directions

due north and due east respectively.

Ex. 2. A point is moving in a straight line with a velocity of 10 feet

per second ;find the component of its velocity in a direction inclined at

an angle of 30 to its direction of motion.

10 TRIANGLE OF VELOCITIES.

19. Components of a velocity in two given directions.

If we wish to find the components of a velocity u in two givendirections making angles a, /3 with it, we proceed as follows.

Let AD represent u in magnitude and direction. Draw AB, ACmaking angles a, /3 with it, and through D draw parallels to complete the

parallelogram ABDC as in Fig. Art. 16. Then we have

AB_ = BD _ ADsin

~~sin a

~sin (a + /3)

'

.-. AB= AD.Binplsin(a-rp),

AC=AD . sin a/sin (a + /3).

Hence the component velocities in these two directions are

u sin /3/sin (a + /3),and wsin a/sin (a + /3).

20. Triangle of Velocities. If a moving point

possess simultaneously velocities represented by the two

sides AB, BC of a triangle taken in order, they are equi-

valent to a velocity represented by AC.

For completing the parallelogram ABCD the lines AB,BC represent the same velocities as AB, AD and hence

have as their resultant the velocity represented by AC.

Cor. 1. If there be simultaneously impressed on a

point three velocities represented by the sides of a triangle

taken in order the point is at rest.

Cor. 2. If a moving point possesses velocities repre-

sented by \ . OA and /A . OB, they are equivalent to a

velocity (X + //,). OG where G is a point on AB such that

For by the triangle of velocities a velocity X . OA is

equivalent to velocities X . OG and X, . GA;also the velocity

yu,. OB is equivalent to

//,. OG and

//-. GB

;but the veloci-

ties \.GA and ft.GB destroy one another; hence the

resultant velocity is (X + /A) . G.

PARALLELOPIPED OF VELOCITIES. 11

21. Parallelepiped of Velocities. By a proof

similar to that for the parallelogram of velocities it maybe shewn that the resultant of three velocities represented

by the three edges of a parallelepiped meeting in a point,

is a velocity represented by the diagonal of the parallele-

piped passing through that angular point. Conversely a

velocity may be resolved into three others.

Similarly, as in Art. 18, it follows that a velocity u maybe resolved into velocities u cos a, u cos /3, u cos 7 along

three directions in space mutually at right angles, where

a, /3, 7 are the angles that the direction of u makes with

these directions.

22. If a moving point possess simultaneously velo-

cities represented by the sides AB, BC, CD,...KL of a

polygon, (whether the sides of the polygon are or are

not in one plane) the resultant velocity is represented

by At.

For, by Art. 20, the velocities AB, BC are represented

by AC; and again the velocities AC, CD by AD and so

on;so that the final velocity is represented by AL.

Cor. If the point L coincides with A (so that the

polygon is a closed figure) the resultant velocity vanishes,

and the point is at rest.

23. When a point possesses simultaneously velocities

in several different directions in the same plane their re-

sultant may be found by resolving the velocities along two

fixed directions at right angles, and then compounding the

resultant velocities.

Suppose a point possesses velocities u, v, w. . . in direc-

tions inclined at angles a, ft, 7,... to a fixed line Ox, and

let Oy be perpendicular to Ox. The components of u

12 UNIFORM MOTION.

along Ox, Oy are respectively wcosa, wsina; the com-

ponents of v are v cos $, v sin /S; and so for the others.

Hence the velocities are equivalent to

u cos a. + v cos yS + w cos 7 parallel to Ox,

and u sin a + w sin 8 + w sin 7 parallel to OT/.

If their resultant be a velocity V at an angle 6 to 0#we must have

V cos = u cos a + v cos /3 +w cos 7 + ,

and V sin Q u sin a + v sin /3 + w sin 7 +

Hence, by squaring and adding,

V2 =(u cos a + v cos /3 +...)" + (u sin a 4- w sin ft + ...)

2

;

, , i ... M sin a -f- t; sin 8 + ...

and, by division, tan = -.

u cos a + v cos p+ . . .

These two equations give J7 and 0.

EXAMPLES.

1. A vessel steams with its bow pointed due north with a velocity of

15 miles an hour, and is carried by a current which flows in a south-easterly

direction at the rate of 3^/2 miles per hour. Find its distance and bearing

from the point from which it started at the end of an hour.

The ship has two velocities one 15 miles per hour northwards and

another 3^/2 miles per hour south-east.

EXAMPLES. 13

Now the latter is equivalent to

3^2 cos 45 or 3 miles per hour eastward,

and 3/2 sin 45 or 3 miles per hour southward.

.-. the total velocity of the ship is 12 miles per hour northwards and

3 miles per hour eastward.

Hence its resultant velocity is v/122+ 32 or ^153 miles per hour in a

direction inclined at an angle tan"11 to the north.

2. A point possesses simultaneously velocities whose measures are 4, 3,

2, 1;the angle between the first and second is 30, between the second and

third 1)0, <ind between the third and fourth 120 ; find their resultant.

Take Ox along the direction of the first velocity and Oy perpendicular

to it.

The angles the velocities make with Ox are respectively 0, 30, 120,

and 240.

Hence, if F be the resultant velocity inclined at an angle 9 to Ox, we

have

F cos = 4 + 3 cos 30 + 2 cos 120 + 1. cos 240;

and . F sin = 3 sin 30 + 2 sin 120 + 1 . sin 240.

Whence we have

Thence

and

Hence the resultant is a velocity ^16 + 9^/3 inclined at an angle

tan-1 (2^/3-

3) to the direction of the first velocity.

3. The velocity of a ship is 8T2T miles per hour, and a ball is bowled

across the ship perpendicular to the direction of the ship with a velocity

of 3 yards per second ; describe the path of the ball in space and shew

that it passes over 45 feet in 3 seconds.

4. A boat is rowed with a velocity of 6 miles per hour straight across

a river which flows at the rate of 2 miles per hour. If its breadth be

300 feet, find how far down the river the boat will reach the opposite

bank, below the point at which it was originally directed.

Ans. 100 feet.

14 EXAMPLES.

5. A ship is steaming in a direction due north across a current

running due west. At the end of one hour it is found that the ship has

made 8^3 miles in a direction 30 west of north. Find the velocity of

the current, and the rate at which the ship is steaming.

Ans. 4^/3 miles per hour; 12 miles per hour.

6. Find the components of a velocity u resolved along two lines

inclined at angles of 30 and 45 respectively to its direction.

Am. (^3-1)1*; U/6 -/$)-.

7. A point which possesses velocities represented by 7, 8, 13 is at

rest ;find the angle between the directions of the two lesser velocities.

Ans. 60.

8. A point possesses velocities represented by 3, 19, 9 inclined at

angles of 120 to one another ;find their resultant.

3 /3Ans. 14 at an L sin"1

-^- with the greatest velocity.

9. A point possesses simultaneously velocities represented by u, 2,

3^/Sw and 4w;the angles between the first and second, the second and

third, and the third and fourth, are respectively 60, 90 and 150;shew

that the resultant is u in a direction inclined at an angle of 120 to that

of the first velocity.

10. A tram-car is moving along a road at the rate of 8 miles per

hour ;in what direction must a body be projected from it with a velocity

of 16 feet per second, so that its resultant motion may be at right angles

to the tram-car ?

Ans. The direction must make an angle cos" 1

(- {\) with the direction

of the car's motion.

11. A ship is sailing north at the rate of 4 feet per second ; the

current is taking it east at the rate of 3 feet per second, and a sailor is

climbing a vertical pole at the rate of 2 feet per second ;find the velocity

and direction of the sailor in space.

12. A point has equal velocities in two given directions ;if one of

these velocities be halved, the angle the resultant makes with the other

is halved also. Shew that the angle between the velocities is 120.

13. A point possesses velocities represented in magnitude and

direction by the lines joining any point on a circle to the ends of a

diameter; shew that their resultant is represented by the diameter

through the point.

ACCELERATION. 15

14. Two steamers A', Y are respectively at two points A, B, 5 miles

apart. A* steams away with a uniform velocity of 10 miles per hour in a

direction making an angle of 60 with AB. Find in what direction Y

must start at the same moment, if it steams with a uniform velocity of

10^/3 miles per hour, in order that it may just come into collision with

X;find also at what angle it will strike X.

Am. At an angle of 150 with AB produced ; it will strike X at right

angles. ^24. Change of Velocity. Suppose a point at any

instant to be moving with a.

velocity represented by OA and

that at some subsequent time

its velocity is represented byOB.

Join AB and complete the

parallelogram OABC.Then velocities represented by OA, OC are equivalent

to the velocity OB. Hence the velocity OC is the

velocity which must be compounded with OA to producethe velocity OB. The velocity 00 is therefore the changeof velocity in the given time.

Thus the change of velocity is not the difference in

magnitude between the magnitudes of the two velocities,

but is that velocity which compounded with the original

velocity gives the final velocity. In the following defini-

tion the expression "change of velocity" is used in this

meaning.

25. Acceleration. Def. The acceleration of a

moving point is the rate of change of its velocity.

The acceleration is uniform when equal changes of

velocity take place in equal intervals of time, however

small these intervals may be.

16 PARALLELOGRAM OF ACCELERATIONS.

When uniform, the acceleration is measured by the

change in the velocity in a unit of time;when variable,

it is measured at any instant by what would be the changeof the velocity in a unit of time, if during that time the

acceleration continued the same as at the instant under

consideration.

26. The definition of the measure of the acceleration at any instant

may also be stated as follows :

Let v be the velocity which must be compounded with the velocity at

the given instant, to produce the velocity which the point possesses at

the time T following the given instant ; then the measure of the accelera-

tion at the given instant is the limit, when r is made indefinitely small,

vof the quantity -

.

27. The unit of acceleration is the acceleration of a

point which moves so that its velocity is changed by the

unit of velocity in each unit of time.

Hence a point is moving with n units of acceleration

when its velocity is changed by n units of velocity in each

unit of time.

Ex. 1. A point is moving eastwards with a velocity of 20 feet per

second, and one hour afterwards it is moving north-east with the same

velocity ;find the change of velocity, and the measure of the acceleration,

supposing the latter to be uniform.

Aits. 20 ^2 -^/2 ft. per second in a direction N.N.W.

Ex. 2. A point is describing a circle of radius 7 yards in 11 seconds,

starting from the end of a fixed diameter, and moving with uniform

speed ; find the change in its velocity after it has described one-sixth of

the circumference.

Ans. 12 ft. -sec. units at an angle of 120 with the initial direction

of motion.

28. Theorem. Parallelogram of Accelerations.

If a moving point have simultaneously two accelerations

1'AKALLELOGKAM OF ACCELERATIONS. 17

represented in magnitude and direction by two sides of a

l><irallelogram drawn from a point, they are equivalent

to an acceleration represented by the diagonal of the

parallelogram passing through that angular point.

Let the accelerations be represented by the sides AB,AC of the parallelogram ABDC, i.e. let AB, AC represent

the velocities added to the velocity of the point in a

unit of time. On the same scale let EF represent the

velocity the particle has at any instant. Draw the

parallelogram EKFL having its sides parallel to AB and

AC; produce EK to M, and EL to N, so that KM, LN

are equal to A B, AC respectively. Complete the parallelo-

grams as in the above figure.

Then the velocity EF is equivalent to velocities EK,EL. But in the unit of time the velocities KM, LN are

the changes of velocity.

/. at the end of a unit of time the component velocities are

equivalent to EM and EN which are equivalent to EO,and this latter velocity is equivalent to velocities EF, FO.

(Art. 20.)

.-. in the unit of time FO is the change of velocity of the

moving point, i.e. FO is the resultant acceleration of the

point.

L. D. 2

18 KELATIVE VELOCITY.

But FO is equal and parallel to A D.

.'.AD represents the acceleration which is equivalent to

the accelerations AB, AC, i.e. is the resultant of the ac-

celerations AB, AC.

29. It follows from the preceding article that ac-

celerations are compounded, and resolved, in the same

manner as velocities are, and propositions similar to those

of Arts. 17 23 will be true if instead of the word "velocity"

we substitute "acceleration".

30. Average Speed and Velocity. The average

speed of a point in a given period of time is the same as

the speed of a moving point which moves uniformly, and

describes the same distance as the given point. Thus the

average speed of a moving point in a given period of

time is the whole distance described by the point in the

given time divided by the whole time. Thus the average

speed of an athlete who runs 100 yards in 101 seconds is

100 -r- 10 or 9-j% yards per second.

The average velocity of a given point in any given

direction (strictly so called) is the whole displacement in

the given direction in the given time divided by the giventime.

31. Relative Velocity. Def. When the distance

between two points is altering, either in direction or in

magnitude or in both, then either point is said to have a

velocity relative to the other, and the relative velocity of one

point with respect to a second is that velocity which, com-

pounded with the velocity of the second, gives the velocity ofthe first point.

Consider the case of two trains moving on parallel rails

in the same direction with equal velocities and let A, B

RELATIVE VELOCITY. 19

be two points, one on each train;a person at one of them,

A say, would, if he kept his attention fixed on B and if

he were unconscious of his own motion, consider B to be

at rest. The line AB would remain constant in magni-tude and direction and the velocity of B relative to Awould be zero.

Next let the first train be moving at the rate of 20

miles per hour and let the second train B be moving at

the rate of 25 miles per hour. In this case the line joiningA to B would be increasing at the rate of 5 miles per hour,

and this would be the velocity of B relative to A.

Thirdly let the second train be moving with a velocity

of 25 miles per hour in the opposite direction to that of

the first;the line joining AB would now be increasing at

the rate of 45 miles per hour in a direction opposite to

that of A's motion, and the relative velocity of B with

respect to A would be 45 miles per hour.

In each of these cases it will be noticed that the

relative velocity of the second train with respect to the

first is obtained by compounding with its own velocity a

velocity equal and opposite to that of the first.

v sin d

Lastly let the first train be moving along the line OCwith velocity u whilst the second train is moving with

velocity v along a line OtD inclined at an angle 6 to OC,

22

20 RELATIVE VELOCITY.

Resolve the velocity v into two components viz. v cos

parallel to OC and v sin 6 in the perpendicular direction.

As before, the velocity of B relative to A, parallel to

OC, would be v cos u; also, since the point A has no

velocity perpendicular to OC, the velocity of B relative to

A in that direction is v sin 6.

Hence the velocity of B relative to A would consist of

two components viz. v cos 9 u parallel to OC and v sin

perpendicular to OC. These two components are equiva-

lent to the original velocity v of the train B together with

a velocity equal and opposite to that of A .

Hence we have the following important result;

If a set of points A, B, C, D,... are in motion, their

relative velocities with regard to any one of them, A say,

are obtained by compounding with their velocities a velocity

equal and opposite to that of A.

32. Rest and motion are relative terms;we do not

know what absolute motion is;all motion that we become

acquainted with is relative.

For example when we say that a train is travelling

northward at the rate of 40 miles an hour we mean that

that is its velocity relative to the earth. Besides this

motion along the surface it partakes with the rest of the

earth in the diurnal motion about the axis of the earth;

it also moves with the earth round the sun;and in

addition has, in common with the whole solar system, any

velocity that that system may have.

33. From the definition in Art. 31 it follows that if

two points A, B are moving in the same direction with

velocities u, v the relative velocity of A with respect to B

RELATIVE VELOCITY. 21

in that direction is u v and that of B with respect to Ais v u.

34. If the two points are not moving in the same

direction the magnitude and direction of the relative

velocity of one with respect to the other may be easily

obtained.

For let PQ, PR represent the velocities u, v of the

points A, B where QPR = <x.

Produce QP to S making PS equal to QP and

complete the parallellogram PSTR.

Then, by Art. 31, PT the resultant of the two velocities

PR, PS represents the relative velocity of B with respect

to A.

Also PT = PR* + RT 2 - '2PR . RT cos a

= u* + v* 2uv cos a.

Also if QPT=0 we have

M _ RT _ smRPT_ sin^-a)v~PR

~suTRTP

~sin B

'

wsinaso that tan 6 = -

.

v cos a u

Hence the relative velocity of B with respect to A is

2 sin ct

Ju* + v* 2uv cos a at an angle tan"1 - with thev cos a u

direction of A.

22 RELATIVE VELOCITY.

35. If the components of the velocity of A parallel

to three fixed directions in space be u, v, w and those of

B be uv v1 ,w

lthen the components of the relative

velocity of B with respect to A parallel to the three fixed

directions are ul u, v^

v and wl

w.

Conversely when the velocity of a point A in space is

known and the velocity of another point B relative to Awe can easily get the velocity of B.

For let u, v, w be the component velocities of Aparallel to three lines fixed in space, u

ltv

t ,w

tthe

corresponding components of B's velocity and U, V, Wthose of the relative velocity of B with respect to A.

Then we have

u1 u=U; .'. u

1= u+ U.

So vt= v + V, and w

l= w+W.

Hence by finding the resultant of u1}

vltw

lwe have

the velocity of B.

36. Relative Accelerations. If we substitute the

word " acceleration"

for"velocity

"in Arts. 31 35, the

results will still be true, since accelerations, like velocities,

follow the parallelogram law.

EXAMPLES.

1. A train is travelling along a horizontal rail at the rate of 60 miles

per hour, and rain is driven by the wind which is in the same direction as

the motion of the train so that it falls with a velocity of 44 feet per second,

and at an angle of 30 with the vertical. Find the apparent direction of

the rain to a person travelling ivith the train.

The true horizontal and vertical components of the rain are 44 sin 30

and 44 cos 30 respectively, or 22 and 22^3 feet per second.

Also the horizontal velocity of train is 60 miles per hour, or 88 feet

per second.

EXAMPLES. 23

/. the components of the velocity of the raiu relative to the train are

- 66 and 22\/3 feet per second.

The resultant of these is a velocity of 44\/3 feet per second, at anfifi

angletan-1 or - 60 with the vertical.

Hence the rain appears to meet the train at an inclination of 60 to

the vertical and the real direction of the rain is at right angles to the

apparent direction.

2. A railway train moving at the rate of 30 miles per hour is struck

by a stone moving horizontally and at right angles to the train with a

velocity of 33 feet per second. Find the magnitude and direction of the

velocity with which the stone appears to meet the train.

Ans. 55 feet per second at an angle tan"1(-

)with the direction

of the train's motion.

3. One ship is sailing south with a velocity of 15^/2 miles per hour,

and another south east at the rate of 15 miles per hour. Find the apparent*'

velocity and direction of motion of the second vessel to an observer on the

first vessel.

Am. 15 miles per hour in a north east direction.

4. In a tunnel, drops of water which are falling from the roof are

noticed to pass the carriage window in a direction making an angletan~* with the horizon, and they are known to have a velocity of 24 feet

per second. Neglecting the resistance of the air find the velocity of the

train. \j

Ans. 32^ miles per hour.

5. A ship is sailing due east, and it is known that the wind is

blowing from the north-west, and the apparent direction of the wind (as

shewn by a vane on the mast of the ship) is from N.N.E. ; shew that the

velocity of the ship is equal to that of the wind.

6. A person travelling eastward at the rate of 4 miles per hour, finds

that the wind seems to blow directly from the north ;on doubling his

speed it appears to come from the north east ; find the direction of the

wind and its velocity.

Ans. 4^/2 miles per hour; towards S.E.

7. A person travelling toward the north east, finds that the wind

appears to blow from the north, but when he doubles his speed it seems

24 ANGULAR VELOCITY.

to come from a direction inclined at an angle cot" 1 2 on the east of north.

Find the true direction of the wind.

Ans. Towards the east.

8. A railway train is moving at the rate of 28 miles per hour, when a

pistol shot strikes it in a direction making an angle sin" 1f with the

train. The shot enters one compartment at the corner furthest from

the engine and passes out at the diagonally opposite corner ;the

compartment being 8 feet long and 6 feet wide, shew that the shot is

moving at the rate of 84 miles per hour nearly, and traverses the carriage

in 7\ths of a second.

9. Two points move simultaneously from points A and B which

are 5 feet apart, one from A towards B with a velocity which would

cause it to reach B in 3 seconds, and the other at right angles to the

former with f of its velocity. Find their relative velocity in magnitude

and direction, the shortest distance between them, and the time when

they are nearest.

Ans. The relative velocity of the 2nd with respect to the 1st is

2T\ ft.-sec. units at an L tan- 1

f with BA ;the shortest distance is 3 feet,

at the end of Iff sees.

10. Two points move with velocities v and 2v respectively in opposite

directions in the circumference of a circle. In what position is their

relative velocity greatest and least and what value has it then ?

11. The radius of the earth being 4000 miles, the latitude X of a point

on the earth's surface at which a train travelling westward at the rate of

9a mile per minute is at rest in space is given by cos X =

'

.

OUTT

37. Angular Velocity. Suppose we have a point

P in motion in a plane, and let be a fixed point in that

plane and Ox a fixed line through 0, then the rate at

which the angle xOP increases is called the angular

velocity of the point P about 0.

If the point move from P to Q in time r and if 6 be

the number of radians in the angle P OQ, then the angulara

velocity of P about is the limit of when T, and

therefore 0, becomes indefinitely small.

ANGULAR VELOCITY. 25

Let V be the linear velocity of the moving point, so

that PQ = V . r (when r is small). Also let OF be the

perpendicular from on the line PQ.

Then

OP . OQ sin 6 = 2 . APOQ = PQ . OF.

Hence

e e sin = Lt PQ OFr

'

OP.OQr sin0- r "an0' r'

OP.OQ" 'OP2

ultimately, when r is very small and p is the perpen-dicular from on the tangent at P to the path of the

moving point.

Hence if to be the angular velocity of the movingpoint about we have

r2 =Vp, where OP = r.

38. The most important case arises when the pointP is describing a circle about as centre. In this case

p = r and we have

Hence, If a moving point describe a circle, its angular

velocity about the centre is equal to its linear velocity divided

by the radius of the circle.

26 ANGULAK VELOCITY.

This important result may be more easily obtained

independently in the case of a point describing a circle.

Let XP be an arc of a circle whose radius is r and

centre 0, and in time T let the point move from P to Q.

Then, since arc PQ = r x

1 arcPg 1 T.= - X- = -. V ,..

r r T r

.'. T(0 = V.

39. The rate of increase of the angular velocity is

called the angular acceleration. It is measured in the

same way as linear acceleration.

40. Areal Velocity. If the path of the moving

point P meet a fixed line Ox in X then the rate of increase

of the area XOP is called the areal velocity of P.

As before, the areal velocity

., area POQ= the limit oi the quantity .

Now area POQ ^PQ x perpendicular from on PQ.

PQ.'. areal velocity

= Lt - x perpendicular fromT

^r'2co. (Art. 37.)

Similarly the rate of increase of the areal velocity is

called the areal acceleration.

EXAMPLES.

1. A wheel rolls uniformly on the ground, without sliding, its centre

describing a straight line; to find the velocity of any point of its rim,

Let be the centre, and r the radius of the wheel, and let v be the

velocity with which the centre advances. Let A be the point of the wheel

in contact with the ground at any instant.

EXAMPLES. 27

Now the wheel turns uniformly round its centre whilst the centre

moves forward in a straight line; also, since each point of the wheel in

succession touches the ground, it follows that any point of the wb.ee*

describes the perimeter of the wheel relative to the centre, whilst the

centre moves through a distance equal to the perimeter; hence the

velocity of any point of the wheel relative to the centre is equal in

magnitude to the velocity v of the centre. Hence any point P of the

wheel possesses two velocities each equal to ?', one along the tangent,

PT, at P to the circle, and the other in the direction in which the centre

is moving.

Hence velocity of A = v - v = 0, and so A is at rest for the instant.

So velocity at B = v + v = 2v.

Consider the motion of any other point P. It has two velocities each

equal to v along PM and PT respectively.

Now the / MPT= L POB= (say).

The resultant of these two velocities v is a velocity 2v cos - along PLm

where / LPT=^=tOPA.m

Hence L APL=OPT-& right angle.

.-. direction of motion of the point P is perpendicular to AP, and its

angular velocity about A

2r cos,

= the angular velocity of the wheel about O.

28 UNIFORMLY ACCELERATED MOTION.

Hence each point of the wheel is turning about the point of contact of

the wheel with the ground, with a constant angular velocity whose measure

is the velocity of the centre of the wheel divided by the radius of the

wheel.

2. A wheel turns about its centre making 200 revolutions per minute ;

what is the angular velocity of any point on the wheel about the centre ?

Ans. 3j-7r.

3. A point moves in a circle with uniform speed ;shew that its

angular velocity about any point on the circumference is constant.

4. A point describes uniformly a given straight line ;shew that its

angular velocity about a fixed point varies inversely as the square of its

distance from the fixed point.

5. A point is describing a parabola with uniform speed ;shew that

its angular velocity about the focus, S, at any point, P, varies inversely as

SP$.

6. A point moves in an ellipse so that its angular velocity about one

focus varies inversely as the square of the radius vector;shew that the

angular velocity about the other focus varies inversely as the square of

the conjugate diameter.

7. An engine is travelling at the rate of 60 miles per hour and its

wheel is 4 feet in diameter ;find the velocity and direction of motion of

each of the two points of the wheel which are at a height of 3 feet above

the ground.

Ans. 88^/3 ft. per sec. at angles 30 with the horizon.

8. If a railway carriage be moving at the rate of 30 miles per hour

and the diameter of its wheel be 3 feet, what is the angular velocity of the

wheel when there is no sliding ;and what is the relative velocity of the

centre and highest point of the wheel ?

Ans. -\8- radians per sec. ;

44 ft. per sec.

41. Theorem. A point moves in a straight line

starting with velocity u, and moving with constant accelera-

tion f in its direction of motion ; if v be its velocity at the

end of time t, when it has moved through a distance s, then

(1) v = u + ft,

(2) s

(3) V2

UNIFORMLY ACCELERATED MOTION. 29

(1) Since f denotes the acceleration i.e. the changein the velocity per unit of time, ft denotes the change in

the velocity in t units of time.

But since the particle possessed u units of velocity

initially, at end of time t it must possess u+ft units, i.e.

v = u+ft.

(2) Again since the velocity increases uniformly

throughout this interval t, the velocity at any time T

preceding the middle of this interval is as much less than

the velocity at the middle of this interval as the velocity

at the time T after the middle of this interval is greaterthan the velocity at the middle of this interval.

Hence the average velocity during the interval must

be the same as that at the middle of the interval, and is

therefore

.'. if s be the space described,

= ut + i/fc*.

(3) The third relation can be easily deduced from

the first two by eliminating t between them.

For from (1), v2 = (u

/. by (2) v* = tf + 2fs.

Cor. If the point move in a curved line, startingwith velocity u, and moving with a constant acceleration

/which at each instant of its motion is directed along the

tangent to its path, the above relations will still be true.


12. Alternative proof of equation (2).

Let the time t be divided into n equal portions of time each equal to

T so that t nr.

Then the velocities of the point at the beginning of each of these

successive intervals are

u, u+fr, M + 2/r, u + (n-l)fr.

Hence the space Sj which loould be moved through by the particle, if it

moved during each of these intervals T with the velocity which it has at

the beginning of each, is

SI= W.T + [+/T] . T+ +[+/(M-I)T].T= nitT+/T

8. {1 + 2 + 3 + (n-l)}

/,' 1\ . t1 I , since r= .

V / n

Also the velocities at the end of each of these intervals are

U +/T, U + 2/T, U + tt/T.

Hence the space s2 which would be moved through by the point if it

moved during each of these intervals r with the velocity it has at the end

of each is

S2= (w+/r) . T + (u + 2/r) . T+ + (w + n/r) . r

= n?ir+/T2(l + 2 + 3 +n)

-}, as before.

Now the true space s is intermediate between s1and s2 ; also the larger

we make n and therefore the smaller the intervals T become, the more

nearly do the two hypotheses approach to coincidence.

If we make n infinitely large the values of slt

s2 both become ut + \f&.

Hence s-ut

43. When the moving point starts from rest we have

u = and the formulae of Art. 41 take the simpler forms

v=ft,

s = W,tf = 2/s.

The proofs of that article with u equated to zero

would apply without other change for these formulae.

UNIFORMLY ACCELERATED MOTION. 31

44. Space described in any particular second.

[The student will notice carefully that the formula (2)

of Art. 41 gives, not the space traversed in the 2thsecond,

but that traversed in t seconds.]

The space described in the t* second

=space described in t seconds space described in (t 1)

seconds

=[ut + W\ -

[u (t-

1) + J/(t-

I)2

]

Hence the spaces described in the first, second, third,

. . .,nth seconds of the motion are

^n lu + /, u + f/, ... u + /

These distances form an arithmetical progression whose

common difference is/!

Hence if a body move with a uniform acceleration the

distances described in successive seconds form an arith-

metical progression whose common difference is equal to

the number of units in the acceleration.

45. The space described in any particular second may be also obtained

by considering the average velocity during that second.

As in Art. 41 the average velocity during the th second is the same

as the velocity at the middle of that second, and is therefore the velocity

at the end of (t-

)seconds from the commencement of the motion.

Hence the average velocity during the tth second

and therefore space described in this second

_8t-l=u+f -- .


EXAMPLES.

1. In what time would a body acquire a velocity of 30 miles per

hour if it started with a velocity of 4 feet per second and moved with the

ft.-sec. unit of acceleration?

Ans. 40 sees.

2. A body starting from rest and moving with uniform acceleration

describes 171 feet in the tenth second ; find its acceleration.

Ans. 18 ft.-sec. units.

3. A point starts with a velocity of 100 cms. per second and moves

with- 2 C.G. s. units of acceleration. When will its velocity be zero, and

how far will it have gone?

Ans. In 50 sees. ; 25 metres.

4. A point moving with uniform acceleration describes in the last

second of its motion ^ ths of the whole distance. If it started from rest,

how long was it in motion and through what distance did it move, if it

described 6 inches in the first second ?

Ans. 5 sees.; 12^ ft.

5. A point moving with uniform acceleration describes 25 feet in the

half second which elapses after the first second of its motion, and 198

feet in the eleventh second of its motion; find the acceleration of the

point and its initial velocity.

Ans. Initial velocity= 30 ft.-sec. units ;acceleration = 16 ft.-sec. units.

6. A point starts from rest and moves with a uniform acceleration of

18 ft. -sec. units;find the time taken by it to traverse the first, second,

and third feet respectively.

/2 1 /3 /2Ans. $, ^-^ , s^- seconds respectively.

o o

7. A point moves over 7 feet in the first second during which it is

observed, and over 11 and 17 feet in the third and sixth seconds respect-

ively; is this consistent with the supposition that it is subject to a

uniform acceleration?

8. A point is moving in a north east direction with a velocity 6, and

has accelerations 8 towards the north and 6 towards the east. Find

its position after the lapse of one second. [The units are a foot and

a second.]

Ans. Its displacement is >/61 + 42^/2 ft. at an angle tan-1

o

north of east.

FALLING BODIES.

When a

34 VERTICAL MOTION UNDER GRAVITY.

But from Art. 43 we know that if a point move from

rest with a constant acceleration the space described is

proportional to the square of the time.

Hence we infer that a falling body moves with a

constant acceleration.

47. From the results of the foregoing and other more

accurate experiments we learn that if a body be let fall

towards the earth in vacua it will move with an acceleration

which is always the same at the same place on the earth,

but which varies slightly for different places.

The value of this acceleration which is called the" acceleration due to gravity

"is always denoted by the

letter"g." The cause of this acceleration will be

discussed in the next chapter.

When foot-second units are used the value of g varies

from about 32'091 at the equator to about 32'252 at the

poles. In the latitude of London its value is about 32'19

and at any point on the earth whose latitude is X is about

32-091 (1 + -005133 sin2

X).

When centimetre-second units are used the extreme

limits are about 978 and 983 respectively, and in the

latitude of London the value is about 981.

The best method of determining the value of "g" is

by means of pendulum experiments ;we shall return to

the subject again in Chapter VIII.

[In all numerical examples unless it is otherwise stated,

the motion may be supposed to be in vacuo and the value

of g taken to be 32 when foot-second units and 981 when

centimetre-second units are used.]

48. Vertical motion under gravity. Suppose a

body is projected vertically from a point qn the earth's

VERTICAL MOTION UNDER GRAVITY. 35

surface so as to start with velocity u. The acceleration of

the body is opposite to the initial direction of motion and

is therefore denoted by g. Hence the velocity of the

body continually gets less and less until it vanishes;the

body is then for an instant at rest but immediately beginsto acquire a velocity in a downward direction and retraces

its steps.

49. Time to a given height. The height h at which

a body has arrived in time t is given by substituting gfor / in equation (2) of Art. 41, and is therefore given by

h = ut -^gf.

This is a quadratic equation with both roots positive ;the

lesser root gives the time at which the body is at the given

height on the way up, and the greater the time at which

it is at the same height on the way down.

Thus the time that elapses before a body which starts with a velocity

of 64 feet per second is at a height of 28 feet is given by

28= 64<-16 2, whence t=\ or .

Hence the particle is at the given height in half a second from the com-

mencement of its motion and again in 3 seconds afterwards.

50. Velocity at a given height and the greatest heightattained.

The velocity v at a given height h is by equation (3)of Art. 41 given by

Hence the velocity at a given height is independent of

the time and is therefore the same at the same pointwhether the body is going upwards or downwards.

32

36 VERTICAL MOTION UNDER GRAVITY.

At the highest point the velocity is just zero; hence

if x is the greatest height attained we have

u*.'. greatest height attained = ~- .

*3

Also the time T to the greatest height is given by= u - gT.

. T- U, . J. .

9

51. Velocity due to a given vertical fallfrom rest.

If a body be dropped from rest its velocity after falling

through a height h is obtained by substituting 0, g, h for

u,f, s in equation (3) Art. 41;

.'. v = J*2gh.

EXAMPLES.

1. A body is projected from the earth vertically with a velocity of

40 feet per second ;find (1) how high it will go before coming to rest,

(2) what times will elapse before it is at a height of 9 feet ?

Am. 25 ft. ; J sec. and 2 sees.

A' 2. A body falls freely from the top of a tower and during the last

second of its flight falls ^| ths of the whole distance. Find the height of

the tower.

Ans. 100 feet.

3. A body moving in a vertical direction passes a point at a height of

54 '5 centimetres with a velocity of 436 centimetres per second;with what

initial velocity was it thrown up, and for how much longer will it rise ?

-4ns. 545 cms. per sec. ; |th sec.

4. A particle passes a given point moving downwards with a velocity

of fifty metres per second; how long before this was it moving upwardsat the same rate ?

Ans. 10-2 sees.

EXAMPLES. 37

6. A body projected vertically downwards described 720 feet in

t seconds, and 2240 feot in 2t seconds; find t and the velocity of

projection.

Ana. t = 5 ; velocity of projection = 64 ft. per sec.

6. A body is projected upwards with a certain velocity, and it is

fonnd that when in its ascent it is 960 feet from the ground it takes

4 seconds to return to the same point again ; find the velocity of projec-

tion and the whole height ascended.

Ana. 256 ft. per sec. ; 1024 ft.

7. A tower is 288 feet high ; ore body is dropped from the top of the

tower and at the same instant another projected vertically upwards from I

the bottom, and they meet half-way up ;find the initial velocity of the

projected body and its velocity when it meets the descending body.

Ant. 96 ft. per sec. ;zero.

8. A body is dropped from the top of a given tower, and at the sameinstant a body is projected, in the same vertical line, with a velocity whichwould be just sufficient to take it to the same height as the tower ; find

where they will meet.

Ans. The first body will have fallen from the top through one

quarter the height of the tower.

9. A body is projected upwards with velocity u, and t seconds

afterwards another body is similarly projected with the same velocity ;

find when, and where, they will meet.

10. A stone is dropped into a well and the sound of the splash is

heard in 7^ seconds ; if the velocity of sound be 1120 feet per second,*

find the depth of the well.

Ant. 784 ft.

11. A stone is dropped into a well and reaches the bottom with a

velocity of 96 feet per second, and the sound of the splash on the water (

reaches the top of the well in 3^ seconds from the time the stone starts ;

find the velocity of sound.

12. A balloon ascends with a uniform acceleration of 4 ft.-sec. units ;

at the end of half a minute a body is released from it ; find the time that

elapses before it reaches the ground.

Ant. 15 sees.

38 MOTION ON SMOOTH

13. A cage in a mine shaft descenda with 2 ft. -sec. units of accelera-

tion. After it has been in motion for 10 seconds, a particle is dropped on

it from the top of the shaft. What time elapses before the particle hits

the cage ?

Ans. 3J sees.

14. Assuming the acceleration of a falling body at the surface of the

moon to be one-sixth of its value on the earth's surface, find the height to

which a particle will rise if it be projected vertically upward from the

surface of the moon with a velocity of 40 feet per second.

Ans. 150 ft.

52. Motion down a smooth inclined plane.

Let AB be a smooth inclined plane inclined at a

given angle a to the horizon and let P be a body on it.

If there were no plane to stop its motion the bodywould fall vertically with an acceleration g.

Now, by the parallelogram of accelerations, a vertical

acceleration g is equivalent to

(1) an acceleration g cos a. perpendicular to the

plane in the direction PR,and (2) an acceleration g sin a down the plane.

The plane prevents any motion perpendicular to itself.

Hence the body moves .down the plane with an

acceleration g sin a, and the investigation of its motion

INCLINED PLANES. 39

is similar to that of a freely falling body, except that

instead of g we have to substitute g sin a in the equation

of Art. 51.

It follows at once that the velocity acquired in sliding

down a length I of the plane is

a .1 = 2g .1 sin a = 2g . PN,

and is therefore the same as that acquired by a particle

in falling freely through a vertical height equal to that

of the plane.

53. If the body be projected up the plane with initial

velocity u an investigation similar to that of Arts. 49 and

50 will give the motion. The greatest distance attained

u2

measured up the plane is _ -.

;the time taken in

2g sin a

traversing this distance is : and so on.

g sin a.

EXAMPLES.

1. A body is projected with a velocity of 80 feet per second up a

smooth inclined plane, whose inclination is 30; find the space described,

and the time that elapses, before it comes to rest.

Am. 200 feet ; 5 sees.

2. A particle slides without friction down an inclined plane and in

the 5th second after starting passes over a distance of 2207'25 centimetres;

lind the inclination of the plane to the horizon.

Am. 30.

3. AB is a vertical diameter of a circle whose plane is vertical, PQ a

diameter inclined at an angle 6 to AB. Find so that the time of sliding

down PQ may be twice that of sliding down AB.

Ans. cos- 1

J.

4. A number of bodies slide from rest down smooth inclined planes

which all commence at the same point and terminate in the same hori-

zontal plane ;shew that the velocities acquired are the same.


5. Two heavy bodies descend the height and length respectively of

an inclined plane ;shew that the times vary as the spaces described and

that the velocities acquired are equal.

6. A heavy particle slides down a smooth inclined plane of given

height ; shew that the time of descent varies as the secant of the inclina-

tion of the plane to the vertical.

54. Theorem. The time that a body takes to slide

down any smooth chord of a vertical circle which is drawn

from the highest point of the circle, is constant.

Let AB be a diameter of a vertical circle of which Ais the highest point and AD any chord.

Let L DAE = 6;and put AD = x, AB = a, so that

x = a cos 0.

As in Art. 52, the acceleration down AD is g cos 9.

Let T be the time from A to D. Then AD is the distance

described in time T by a particle smarting from rest and

moving with acceleration g cos 6.

T = / 2x = I^LV gcos0 V g

LINES OF QUICKEST DESCENT. 41

This result is independent of 6 and is the same as the

time of falling vertically through the distance AB.

Hence the time of falling down any chord of this

circle beginning at A is the same.

55. The same theorem will be found to be true for

all chords of the same circle ending in the lowest point ;

or if the plane of the circle be inclined to the vertical.

The theorem is also true if we substitute, for the

circle, a sphere having A as the highest point; for any

plane section of a sphere is a circle and hence the

theorem is true for all plane sections of the sphere

through the vertical diameter, and hence for the sphere.

EXAMPLES.

1. If two circles touch each other at their highest or lowest points

and a straight line be drawn through this point to meet both circles, then

the time of sliding from rest down the portion of this line intercepted

between the two circles is constant.

2. A body slides down chords of a vertical circle ending in its lowest

point ; shew that the velocity on reaching the lowest point varies as the

length of the chord.

56. Lines of quickest descent. The line of

quickest descent from a given point to a curve in the

same vertical plane is the straight line down which a

body would slide from the given point to the given curve

in the shortest time. It is not the same line as the

geometrically shortest line that can be drawn from the

given point to the curve. For example, the straight line

down which the time from a given point to a given planeis least, is not, in general, the perpendicular from the given

point upon the given plane.

42 LINES OF QUICKEST DESCENT.

57. Theorem. The chord of quickest descentfrom a

given point P to a curve in the same vertical plane is PQ,where Q is a point on the curve such that a circle having Pas its highest point touches the curve at Q.

For let a circle be drawn having its highest point at

P to touch the given curve in Q. Take any other point

Q! on the curve and let PQ tmeet the circle again in R.

Then since P^ is > PR.

/. time down P Qlis > time down PR.

But time down PR = time down PQ (Art. 54).

.'. time down PQ lis > time down PQ,

and Q1is any point on the given curve.

.'. time down PQ is less than that down any other straight

line from P to the given curve.

It is clear by the construction that the chord of

quickest descent makes equal angles with the vertical

and the normal to the curve.

Cor. I. If we want the chord of quickest descent from

a given curve to a given point P, we must describe

EXAMPLES. 43

a circle having the given point P as its lowest point to

touch the curve in Q ;then QP is the required straight

line.

Cor. II. If instead of a curve we have to find the

chord of quickest descent from a point to a given surface,

we must describe a sphere having the given point as

highest point and touching the given surface;then the

straight line from the given point to the point of contact

of the sphere is the chord required.

EXAMPLES.

1. To find the straight line of quickest descent to a given straight line

from a given point P in the same vertical plane.

Let EG be the given straight line. Then we have to describe a circle

having its highest point at P to touch the given straight line. Draw PBhorizontal to meet DC in B. From BC cut off a portion BQ equal to BP.Then PQ is the required chord ; for it is clear that a circle can be drawnto touch BP and BQ at P and Q respectively.

2. To find the line of quickest descent from a given point to a givencircle in the same vertical plane.

Join P to the lowest point B of the given circle to meet the circle

44 LINES OF QUICKEST DESCENT.

again in Q. Then PQ is the required line. For join 0, the centre of the

circle, to Q and produce to meet the vertical through P in G.

The QPC=WBQ, since OB and CP are parallel,

:. a circle whose centre is C, and radius CP, will have its highest point at

P and will touch the given circle at Q.

If P be within the given circle, join P to the highest point and produceto meet the circumference in Q ; then PQ will be the required line.

'' 3. To shew that the line of quickest descent from one curve to another

in the same vertical plane makes the same angle with the normals to the

curves at the points where it meets them.

For let PQ be the chord of shortest descent from one curve to the

other.

EXAMPLES. 45

Let OQ, OjP be the normals at Q, P and OP, OjQ the verticals.

Then since of all chords drawn through P, PQ is the line of quickest

descent from P to the lower curve

:.tOQP=LOPQ. (Art. 57.)

So since of all chords drawn from the upper curve to end in Q, PQ is

the line of quickest descent

But since PO, QO^ are parallel

Hence LOQP=L01PQ,

and hence the proposition is true.

It may be noticed that the line of quickest descent is the diagonal of a

rhombus formed by the normals and the vertical lines drawn through its

two ends.

4. An ellipse is placed with its minor axis vertical; shew that the

normal chord of quickest descent from the curve to the major axis is that

drawn from a point at which the foci subtend a right angle, if there is such

a point. What is the condition that there may be such a point, and whatis the normal chord of quickest descent if there is not ?

Let the tangent and normal at P meet the minor axis in t and g.

Then since the ratio of PG to Pg is the same for all positions of P, the

time down Pg is a minimum also. But since tPg is a right angle, the

time down Pg is the same as the time down tg.

:. the time down tg, and hence the length tg, is a minimum.

46 GRAPHIC METHODS.

But tg is the diameter of a circle passing through S, P, and H ; and

the least circle passing through S, H has its centre at G.

Hence we must have SPH a right angle.

Again since lSPH=s.StH, it is less than LSBH.

.'.there is no such point unless LSBH is >^ i.e. unless >sin- or2 a 4

If e be < p the minor axis BG will be the required line.

V 2

58. Graphic Method. Velocity-Time Curve.To determine the distance described in a given time when

the velocity of the moving point is varying.

Take two straight lines Ox, Oy at right angles to one

another and let times be represented by lengths drawn

along Ox so that a unit of length in this direction

represents a unit of time.

Let BG be a line, curved or straight, such that an

ordinate to it represents the velocity of the moving pointat the time which is represented by the correspondingabscissa. Thus if MP be an ordinate to the curve, it

represents the velocity of the moving point at a time

VELOCITY-TIME CURVE. 47

from the commencement of the motion represented byOM.

We shall shew that the space described by the moving

point in time OA is represented by the area bounded by

OB, OA,AC and the curved line BC.

Take an ordinate QN close to PM. Then during the

time MN the point moves with a velocity greater than

MP and less than NQ. Hence the number of units of

space it describes is intermediate between the number of

units of area in the rectangles PN and QM. Similarly if

we divide OA into any number of parts and erect paral-

lelograms on each.

Hence the total space described in time OA is inter-

mediate between the space represented by the sum of

the inner rectangles and that represented by the sum of

the outer rectangles. Now let the number of portions

into which the time OA is divided be increased inde-

finitely in which case the area of the inner series of

rectangles and that of the outer series become both

ultimately equal to the area of the curve.

[For the difference between the areas of the inner and outer series of

rectangles is less than the area of a rectangle whose height is AC and

whose base is the greatest of the parts into which OA is divided, and

therefore vanishes ultimately when the parts into which OA is divided

are taken indefinitely small. Hence the areas of the two series of rect-

angles are ultimately equal and therefore the area of the curve, which

necessarily lies between them, must ultimately be equal to either.]

Hence the number of units of space described by the

moving point is ultimately equal to the number of units

of area in the area OACB.

Cor. Since RQ is the increase of velocity in time

MN, the acceleration of the moving point at this instant

48 GRAPHIC METHODS.

is equal to the limit of j^. when MN is made indefinitely

small.

But when MN is made indefinitely small, the point Q

moves up to P, PQ becomes the tangent at P and

is ultimately the tangent of the angle that the tangent at

P makes with Ox.

Hence in the velocity-time curve the numerical value

of the acceleration is the slope of the curve to the line

along which the time is measured.

59. Case of uniform acceleration. Here since the

acceleration is constant the slope is constant, and the

velocity-time curve therefore consists of a straight line

inclined at a constant angle to the time-line.

[This can be easily shown ab initio.

velocity at any time t = u +ft we have

MP = OB +f. OM.

For since the

/=MP-OBOM

= tan TBP,

so that TBP is a constant angle.]

ACCELERATION-TIME CURVE. 49

Hence the number of units of space described in time t

= the number of units of area in OACBu + u -

nA (OB + AC\ _ [*YT2 ~; u

= ut +

60. Acceleration-Time Curve. If the ordinate to

the curve drawn in Art. 58 represent the acceleration of

the moving point [so that MP represents the acceleration

of the moving point at a time from the commencement of

the motion represented by OM], then by a method of

reasoning similar to that article, the velocity acquired bythe moving point at the end of time OA is represented

by the number of units of area in OACB.

EXAMPLES.

1 . Draw a diagram in the case when a particle starts from rest and

moves so that the product of its velocity and acceleration is constant;

and find the space described in a given time.

The velocity-time curve in this case is such that its subnormal is

constant and hence is a parabola. The space described in time t is

f A/2\t*, where X is the given constant.

2. If fee t then the velocity at any time oc t2 and the space described

oc3.

3. If/oc t2 then the velocity oc t

3 and the space described <z t4

.

4. If at any time the velocity of a moving point be proportional to

the square of the time that has elapsed from the commencement of its

motion, shew that the acceleration at any time is equal to twice the ratio

of the velocity to the time that has elapsed and that the space described

is one-third the product of the velocity and the time.

L. D.

50 EXAMPLES.

EXAMPLES. CHAPTER I.

1. A particle is dropped -from a height h, and after

falling rds of that distance passes a particle which was

projected upwards at the instant when the first was dropped.

Find to what height the latter will attain.

2. A body begins to slide down a smooth inclined planefrom rest at the top, and at the same instant another body is

projected upwards from the foot of the plane with such a

velocity that they meet half way up the plane ;find the

velocity of projection and determine the velocity of each when

they meet.

3. Assuming that the earth in a year describes a circle

uniformly about the sun as centre, that the distance between

the centres is 240 radii of the sun and that the radius of the

sun is 100 times that of the earth, find the velocity of the

vertex of the earth's shadow taking the sun's radius as the

unit of space and a year as that of time.

4. A body starts with velocity u and moves with uniform

acceleration;

if a, b, c are the spaces described in the pth, qth

and rth seconds respectively, shew that

a (q-

r) + b (r-p) + c (p

-q)= 0.

5. If a space s be divided into n equal parts at the end

of each of which the acceleration of a moving point is increased

/by

-,find the velocity of the point after describing the distance

iff

s if it started with velocity u.

EXAMPLES ON CHAPTER I. 51

6. A particle starts from rest with acceleration f\ at

the end of time t the acceleration becomes 2f; 3/ at end of

time 2, and so on. Find the velocity at the end of time nt

and shew that the space described is

7. If the acceleration of a particle for the first second be

a and if it increase at the end of each second in geometrical

progression with a common ratio r, shew that the space de-

scribed in n seconds is

rn - In + I rn - r

8. If a particle occupy n seconds less and acquire a

velocity of m feet per second more at one place than at

another in falling through the same distance;shew that -

72-

equals the geometrical mean between the numerical values of

gravity at the two places.

9. A train goes from rest at one station to rest at another,

one mile off, being uniformly accelerated for the first frds of

the journey and uniformly retarded for the remainder, and

takes 3 minutes to describe the whole distance. Find the

acceleration, the retardation, and the maximum velocity.

10. An engine driver suddenly puts on his brake and

shuts off' steam when he is running at full speed ;in the first

second afterwards the train travels 87 feet, and in the next

85 feet. Find the original speed of the train, the time that

elapses before it comes to rest, and the distance it will travel

in this interval, assuming the brake to cause a constant re-

tardation. Find also the time the train will take, if it is 96

yards long, to pass a spectator standing at a point 484 yardsahead of the train at the instant when the brake was applied.

42

52 EXAMPLES ON CHAPTER I.

11. Two points describe the same circle in such a manner

that the line joining them always passes through a fixed point;

shew that their velocities at any instant are proportional to

their distances from the fixed point.

12. A cannon ball is moving in a direction making an

angle 9 with a line drawn from the ball to an observer;

if Fbe the velocity of sound and nV that of the ball, shew that

the whizzing of the ball will be heard in the order in which it

is produced or in the reverse order according as n is ^ sec 9.

13. Two particles P, Q start simultaneously from a point

A, one sliding down a plane AS inclined at an angle a to the

horizon, and the other falling freely; shew that their relative

acceleration is g cos a.

Hence shew that the line PQ is always perpendicular to AB.

14. Find the relative motion of two points one of which

revolves in a circle of radius a with angular velocity CD, and

the other moves with velocity aw in a straight line passing

through the centre of the circle.

15. Two points describe concentric circles uniformly, the

time of describing the outer circle being m times that in the

inner;

v is the speed in the former and u in the latter ;

shew that when the angular velocity of the one relative to the

other is zero the actual velocity of the one relative to the

other is

Im - 1 ,

\/ i (u v

)-V m + l^

16. Two points describe concentric circles of radii a and

b with velocities varying inversely as the radii;shew that the

relative velocity is parallel to the line joining the points when

the angle between the radii to these points is


17. If the distance between two moving points at any

time be a, V their relative velocity and u, v the components

of V in and perpendicular to the direction of a, shew that

wotlirir distance when they are nearest to one another is -==

,and

ctuthat the time before they arrive at their nearest distance is-f

18. Shew that the highest point of a wheel rolling on a

horizontal plane moves twice as fast as a point on the rim

whose distance from the ground is half the radius.

19. Two particles are describing in the same sense equal

circles in the same plane with the same constant speed.

Give a geometrical construction to find the position of the

particles in which each appears to the other to be stationary,

and shew that the intervals between successive stationary

points are in the ratio TT + a 20 : ir a + 20, where a is the

radius of either circle, 2c the distance between their centres, a

the angle between the central radii of the particles, and is

given by the equation

csn - = -.

20. A body at the equator falls to the earth from relative

rest. Shew that it will fall to the east of the vertical line in

which the fall began, and find approximately the amount of

the deviation in terms of h, the vertical height through which

it has fallen.

21. If be the circumcentre and Olthe orthocentre of a

triangle ABC, then velocities represented by OA, OB, OC are

equivalent to 00^, and velocities O^A, 0^, }C to one

represented by twice Ofi.

22. A particle has simultaneously impressed on it three

velocities represented by PA, PB, PC;shew that it will move

in the direction PQ where Q is the centroid of the triangle

ABC.

54 EXAMPLES ON CHAPTER I.

23. AO, BO are diameters of two circles which are in a

vertical plane and touch at ; shew that if POQ be drawn

through bisecting the angle between AOB and the vertical

through and meeting the circles in P and Q, then the time

down POQ is independent of the inclination of AOB to the

vertical; also if AOB be vertical the time down any chord

POQ is constant.

24. Bodies slide down smooth faces of a pyramid starting

from rest at the vertex; shew that at any time t they all lie

on a sphere whose radius is \gfi.

25. A parabola is placed with its axis vertical and vertex

upwards; find (1) the chord of quickest descent from the focus

to the curve, (2) the focal chord of quickest descent.

26. A parabola is placed in a vertical plane with its axis

inclined to the vertical. S is the focus, A the vertex, and Qthe point of the curve vertically below S; if SP be the straight

line of quickest descent from the focus to the curve, shew that

L ASP = twice L PSQ.

27. A parabola is placed with its axis horizontal; find the

normal chord of quickest descent from the curve to the axis,

and the chord of quickest descent from the focus to the curve.

28. An ellipse is placed with its major axis vertical;

shew that the line of quickest descent from its upper focus to

the curve is equal to the length of the latus rectum provided

the eccentricity of the ellipse is > .

29. An ellipse is placed with its plane vertical and minor

axis horizontal; find the line of quickest descent from the

curve to its centre and determine its length.

30. The time of descent from rest down chords of an

ellipse passing through its lowest point is a maximum or a

minimum for those chords which are parallel to the transverse

and conjugate axes respectively.


31. An ellipse is placed with its major axis vertical;

shew that the time down any chord of its auxiliary circle

which touches the ellipse is constant.

32. A hyperbola is placed in a vertical plane with its

transverse axis horizontal;shew that the time of descent

down a diameter is least when the conjugate diameter is equal

to the distance between the foci.

33. Shew that the time of quickest descent from a given

yC)J

f\

sec x ,where is the

*7

inclination of the straight line to the horizontal, and I is the

shortest distance between the straight line and the circle.

34. Shew that the time of quickest descent from one circle

to another in the same vertical plane is . / ~-r- .V ga + b + c sin a

where a, b are the radii, c the distance between the centres,

and a the inclination to the horizon of the line joining the

centres.

35. Two parabolas are placed with axes vertical, vertices

downwards, and foci coincident. Shew that there are three

chords down which the time of descent of a particle under

gravity is a minimum, one being the principal diameter and

the other two making angles of 60 on either side of the

vertical.

36. The locus of the points from which the time of

descent to two given points is the same is a rectangular

hyperbola.

37. Find the locus of a point from which the time of

quickest descent to a given vertical circle is the same.

CHAPTER II.

THE LAWS OF MOTION.

61. IN the last chapter we discussed a few simple

cases of motion without any reference to the way in which

motion is produced. In the present chapter we propose to

consider the production of motion, and it will be necessary

to commence with a few elementary definitions.

Matter is "that which can be perceived by the senses"

or "that which can be acted upon by, or can exert, force."

No definition can however be given that would convey

an idea of what matter is to anyone who did not already

possess that idea. It, like time and space, is an elementary

conception.

A Body is a portion of matter which is bounded bysurfaces and which is limited in every direction.

A Particle is a portion of matter which is infinitely

small in all its dimensions.

1 The Mass of a body is the quantity of matter in the

body.

Force is that which changes, or tends to change, the

state of rest or uniform motion of a body.

THE LAWS OF MOTION. 57

62. These definitions may appear to the student to be

vague, but we may illustrate their meaning somewhat as

follows.

If we have a small portion of any substance, say iron,

resting on a smooth table we may by a push be able to

move it fairly easily ;if we take a larger quantity of the

same iron the same effort on our part will be able to move

it less easily. Again, if we take two portions of platinumand wood of exactly the same size and shape the effect

produced on these two substances by equal efforts on our

part will be quite different. Thus common experienceshews us that the same effort applied to different bodies

under seemingly the same conditions does not always pro-

duce the same result. This is because the masses of the

bodies are different.

63. If to the same mass we apply two forces in

succession, and they generate the same velocity in the

same time, the forces are said to be equal.

If the same force be applied to two different masses and

if it produce in them the same velocity in the same time,

the masses are said to be equal.

The student will notice that we here assume that it is

possible to create forces of equal intensity on different

occasions, e.g. we assume that the force necessary to keepa spiral spring stretched through the same distance is

always the same when other conditions are unaltered.

Hence by applying the same force in succession we can

obtain a number of masses each equal to a standard unit

of mass.

64. The British unit of mass is called the Imperial

Pound, and consists of a lump of platinum deposited in the

58 THE LAWS OF MOTION.

Exchequer Office, of which there are in addition several

accurate copies kept in other places of safety.

The French or scientific unit of mass is called a

gramme, and is the one-thousandth part of a certain

quantity of platinum deposited in the Archives. The

gramme was meant to be defined as the mass of a cubic

centimetre of pure water at a temperature of 4 C. It is

a much smaller unit than a Pound.

One Gramme = about 15 '432 grains.

One Pound = about 453'6 grammes.

The system of units in which a centimetre, gramme,and second, are respectively the units of length, mass and

time, is generally called the C. G. s. system of units.

65. Density. The density of a body when uniform

is the mass of a unit volume of the body ;so that if m

be the mass of volume V of a body whose density is p, then

m= Vp.

When variable, the density at any point of a body is the

ratio of the mass of a very small portion of the body sur-

rounding the point to the volume of that portion, so that

Tfl

p = Limit of-p.,

when V approaches to zero.

66. The Weight of a body is the force with which

the earth attracts the body.

It can be shewn that every particle of matter in nature attracts every

other particle with a force which varies directly as the product of the

masses of the quantities and inversely as the square of the distance

between them;hence it can be deduced that a sphere attracts a particle

on, or outside, its surface with a force which varies inversely as the square

of the distance of the particle from the centre of the sphere. The earth

DEFINITIONS. 59

is not accurately a sphere and therefore points on its surface are not

equidistant from the centre ;hence the attraction of the earth for a given

mass is not the same at all points of its surface, and therefore the weight

of a given mass is different at different points of the earth.

67. The Momentum or Quantity of Motion of a

body is proportional to the product of the mass and

the velocity of the body.

If we take as the unit of momentum the momentum of

a unit mass moving with unit velocity, then the momentum

of a body is mv, where m is the mass and v the velocity of

the body. The direction of the momentum is the same as

that of the velocity.

68. The Vis Viva and Kinetic Energy of a bodyare both proportional to the product of the mass and the

square of the velocity of the body. If the unit of vis

viva be that of a unit mass moving with unit velocity

then the vis viva of a body is mv2.

It is convenient to define the kinetic energy of a bodyas %-mv

1

,so that it is half its vis viva. It will be noted that

the unit of kinetic energy is that of a mass equal to twice

the unit of mass moving with unit velocity.

The kinetic energy of a body has magnitude only,

and not direction. [Such a quantity is called a Scalar

quantity. A quantity which has magnitude and direction,

such as a velocity or an acceleration, is called a Vector

quantity.]

69. We can now enunciate what are commonly called

Newton's Laws of Motion. " The first two were discovered

by Galileo, and the third in some of its many forms was

known to Hooke, Huyghens, Wallis, Wren and others

before the publication of the Principia."


The Laws of Motion are;

Law I. Every body continues in its state of rest,

or of uniform motion in a straight line, except in so faras it be compelled by impressed force to change that

state.

Law II. The rate of change of momentum is pro-

portional to the impressed force, and takes place in tie

direction of the straight line in which the force acts.

Law III. To every action there is an equal and

opposite reaction.

No formal proof, experimental or otherwise, can be givenof these three laws. On them however is based the whole

system of Dynamics, and on Dynamics the whole theory of

Astronomy. Now the results obtained, and the predictions

made, from the theory of Astronomy agree so well with the

actual observed facts of Astronomy that it is inconceivable

that the original laws on which the subject is based should

be erroneous. For example, the Nautical Almanac is pub-lished four years beforehand and the predictions in it are

always correct. Hence we believe in the truth of the above

three laws of motion, because the conclusions drawn from

them agree with our experience.

70. Law I. We never see this law practically

exemplified in nature, because it is impossible ever to getrid of all forces during the motion of the body. It maybe seen approximately in operation in the case of a pieceof dry hard ice projected along the surface of dry, well

swept, ice. The only forces acting on the fragment of ice,

in the direction of its motion, are the friction between the

two portions of ice and the resistance of the air. The

LAW I. 61

smoother the surface of the ice the further the small

portion will go, and the less the resistance of the air the

further it will go. The above law asserts that if the ice

were perfectly smooth and if there were no resistance of

the air and no other forces acting on the body, then it would

go on for ever in a straight line with uniform velocity.

The law states a principle sometimes called the Prin-

ciple of Inertia, viz. that a body has no innate tendencyto change its state of rest or of uniform motion in a straight

line. If a portion of metal attached to a piece of string be

swung round on a smooth horizontal table, then, if the

string break, the metal, having no longer any force actingon it, proceeds to move in a straight line, viz. the tangentto the circle at the point at which its circular motion

ceased.

If a man step out of a rapidly moving train he is

generally thrown to the ground ;his feet on touching the

ground are brought to rest; but, as no force acts on the

upper part of his body, it continues its motion as before,

and the man falls to the ground.If a man be riding on a horse which is galloping at a

fairly rapid pace and the horse suddenly stop, the rider is

in danger of being thrown over the horse's head.

The above examples give good illustrations of the law

and tend to elucidate its meaning.We deduce from the law that if a body be moving

uniformly in a straight line the total force acting on the

body must be zero, and hence the forces, if any, which act

on the body must balance one another.

71. Law II. From this law we derive our method

of measuring force.

62 LAW II.

Let m be the mass of a body, and / the acceleration

produced in it by the action of a force whose measure

is P.

Then by the second law of motion

P <x rate of change of momentum,oc rate of change of mv,

<x m x rate of change of v (m being unaltered),

ccm.f.

.-. P =\.rnf, where \ is some constant.

Now let the unit of force be so chosen that it may

produce in unit mass the unit of acceleration.

/. when m = 1 and /= 1, we have P = 1,

and therefore X = 1.

Hence, in this case, we have

72. Magnitude of this unit of force. As explained

in Art. 47 we know that when a body drops vertically in

vacua it moves with an acceleration which we denote by

"g" ',

also the force which causes this acceleration is what

we call its weight.

Hence

The weight of the unit of mass acting on the unit mass

produces g units of acceleration;

.'.-

. weight of unit mass acting on the unit mass producest7

one unit of acceleration [by the second law].

But the unit of force is that which produces unit

acceleration in unit mass;

. the unit of force = x weight of unit mass.9

FUNDAMENTAL EQUATION. 03

73. In our ordinary system of units the unit of mass

is one pound and g = 32'2 approximately. Hence the unit

of force is equal to ^s^n times the weight of a pound, andoZ'2i

is therefore equal to the weight of about half an ounce.

This unit is the British Absolute Unit of Force and is

called a Poundal, so that

g . Poundals = weight of a lb.,

where g = 32'2 approximately.

Hence the equation P = mf is a true relation,m being the number of pounds in the body, P the

number of poundals in the force acting on it, and f

the number of units of acceleration produced in

the mass m by the action of the force P on it.

This relation is sometimes expressed in the form

Moving ForceAcceleration = ,,

Mass moved

N.B. All through this book the unit of force used will

be a poundal unless it is otherwise stated. Thus when we

say that the tension of a string is T we mean T poundals,

Tso that the tension is equal to the weight of pounds.

\J

74. C. G. S. system of units. In this system the

unit of mass is a gramme, and g = 981 approximately.The unit of force in this system is therefore equal to -^of the weight of a gramme. This unit is the C. G. s.

absolute unit of force and is called a Dyne ;so that,

g . Dynes = weight of one gramme,where g = 981 approximately.

Also One Poundal = about 13800 Dynes.

64 THE WEIGHT OF A BODY

Hence in the c. G. s. system when we use the equation

P mf, the force must be expressed in dynes, and the

mass in grammes.

75. A poundal and a dyne are called Absolute Units

because their values are not dependent on the value of g,

which varies at different places on the earth's surface. The

weight of a pound and of a gramme do depend on this

value. Hence they are called Gravitation Units.

76. The weight of a body is proportional to its mass

and is independent of the kind of matter of which it is com-

posed. The following is an experimental fact : If we have

an air-tight receiver, and if we allow to drop at the same

instant, from the same height, portions of matter of anykind whatever, such as a piece of metal, a feather, a piece

of paper etc., all these substances will be found to have

always fallen through the same distance, and to hit the base

of the receiver at the same time, whatever be the sub-

stances, or the height from which they are allowed to fall.

Since these bodies always fall through the same height

in the same time, therefore their velocities [rates of change

of space,] and their accelerations [rates of change of velo-

city,] must be always the same.

Let Tfj, TF2 poundals be the weights of any two of

these bodies, mv m2their masses. Then since their ac-

celerations are the same and equal to g, we have

.'. Wl

: W2

: : m1

: m2 ,

or the weight of a body is proportional to its mass.

Hence bodies whose weights are equal have equal

IS PROPORTIONAL TO ITS MASS. 65

in;usses; so also the ratio of the masses of two bodies

is known when the ratio of their weights is known.

N.B. The equation Wmg is a numerical one and

means that the number of units of force in the weight of

a body is equal to the product of the number of units of

mass in the mass of the body, and the number of units of

acceleration produced in the body by its weight.

77. Distinction between mass and weight. The student

must carefully notice the difference between the mass and

the weight of a body. He has probably been so accustomed

to estimate the masses of bodies by means of their weightsthat he has not clearly distinguished between the two. If

it were possible to have a cannon ball at the centre of the

earth it would have no weight there;for the attraction of

the earth on a particle at its centre is zero. If however it

were in motion the same force would be required to stop

it as would be necessary under similar conditions at the

surface of the earth. Hence we see that it might be

possible for a body to have no weight ;its mass however

remains unaltered.

^The confusion is probably to a great extent caused by

the fact that the word "pound" is used in two senses

which are scientifically different;

it is used to denote both

what we more properly call "the mass of one pound" and

"the weight of one pound." It cannot be too strongly

impressed on the student that, strictly speaking, a pound is

a mass and a mass only ;when we wish to speak of the

force with which the earth attracts this mass we ought to

speak of the "weight of a pound." This latter phrase is

often shortened into "a pound," but care must be taken to

see in which sense this word is used.

L. D.


78. Weighing by Scales and a Spring Balance. Wehave pointed out (Art. 47) that the acceleration due to

gravity, i.e. the value of g, varies slightly as we proceed

from point to point of the earth's surface. When we

weigh out a substance (say tea) by means of a pair of

scales, we adjust the tea until the weight of the tea is

the same as the weight of sundry pieces of metal whose

masses are known, and then, by Art. 76, we know that

the mass of the tea is the same as the mass of the metal.

Hence a pair of scales really measures masses and not

weights, and so the apparent weight of the tea is the same

everywhere.When we use a spring balance however we compare the

weight of the tea with the force necessary to keep the

spring stretched through a certain distance. If then we

move our tea and spring balance to another place, say

from London to Paris, the weight of the tea will be differ-

ent whilst the force necessary to keep the spring stretched

through the same distance as before will be the

Hence the weight of the tea will pull the spring thrc

a distance different from the former distance, and hence

its apparent weight as shewn by the instrument will be

different.

If we have two places, A and B, at the first of which

the numerical value of g is greater than at the second,

then a given mass of tea will [as tested by the spring

balance] appear to weigh more at A than it does at B.

At the centre of the earth no mass has any weight at

all, so that there we could not compare the masses of bodies

by means of a spring balance.

EXAMPLES. 67

EXAMPLES.

1. A mass of 10 pounds is placed on a smooth horizontal plane, andis acted on by a force equal to the weight of 3 pounds ;

find the distance

described by it in 10 seconds.

Here moving force= weight of 3 Ibs. = 3# poundals ;

and mass moved= 10 pounds.

Hence, using ft.-sec. units, and supposing = 32, the acceleration =~ .

.: distance required= .~

. 102=480 feet.

2. Find the magnitude of the force which, acting on a kilogrammefor 5 seconds, produces in it a velocity of one metre per second.

Here velocity acquired = 100 cms. per sec.;

.-. acceleration = 20 c. G. s. units.

/. Force = 1000 x 20 dynes = weight of about or 20-4 grammes./oi

3. Find the magnitude of the force which acting on a mass of

10 cwt. for 10 seconds will generate in it a velocity of 3 miles per hour,

Am. Wt. of 15? Ibs.

4. A pressure equal to the weight of a kilogramme acts on a body

continuously for 10 seconds and causes it to describe 10 metres in that

time ; find the mass of the body.

Ans. 49'05 kilogrammes.

5. A horizontal pressure equal to the weight of 9 pounds acts on a

mass along a smooth horizontal plane ; after moving through a space of

25 feet, the mass has acquired a velocity of 10 feet per second;find its

magnitude.

Ans. 144 Ibs.

6. A body is placed on a smooth table and a pressure equal to the

weight of 6 pounds acts continuously on it ; at the end of three seconds

the body is moving at the rate of 48 feet per second ; find the mass of

the body.

Ans. 12 Ibs.

7. A body of mass 200 tons is acted on by a force equal to 112000

poundals ;how long will it take to acquire a velocity of 30 miles per hour ?

Ans. 2 min. 56 sees.

8. In what time will a force equal to the weight of 10 Ibs. acting on a

mass of one ton move it through 14 feet ?

Ans. 14 sees.

52

68 EXAMPLES.

N9. A mass of 10 Ibs. falls 10 feet from rest, and is then brought to

rest by penetrating 1 foot into some sand ;find the average pressure of

the sand on it.

Ans. Weight of 100 Ibs.

10. A certain force acting on a mass of 10 pounds for 5 seconds pro-

duces in it a velocity of 100 feet per second. Compare the force with the

weight of a pound, and find the acceleration it would produce in a ton.

Ans. 6J : 1; fa ft.-sec. units.

11. A train of mass 200 tons is^urged forward[by a force equal to the

weight of 1^ tons while the resistance it experiences is equal to the

weight of one ton. How long will it take to acquire a velocity of 10 miles

per hour ?

Ans. 3 min. 3^ sees.

12. A bullet moving at the rate of 200 feet per second is fired into

a trunk of wood into which it penetrates nine inches;

if a bullet moving

with a similar velocity were fired into a similar piece of wood five inches

thick, with what velocity would it emerge supposing the resistance to be

uniform ?

Ans. 133 ft.-sec. units.

13. A cannon ball of mass 1000 grammes is discharged with a

velocity of 45000 centimetres per second from a cannon the length of

whose barrel is 200 centimetres ;shew that the mean force exerted on the

ball during the explosion is 5-0625 x 109dynes.

14. At the equator the value of g is 32-09 and in London = 32-2 ; a

merchant buys tea at the equator, at a shilling per pound, and sells in

London; at what price per Ib. (apparent) must he sell so that he may

neither gain nor lose, if he uses the same spring balance for both

transactions ?

A quantity of tea which weighs one Ib. at the equator will appear to

qo.o 32*2weigh ^--jrj: Ibs. in London. Hence he should sell n^-rn Ibs. for one

' *

3209shilling, or at -^ shillings per Ib.

15. At a place A, = 32-24 and at another place JB, gr=32-12. Amerchant buys goods at 10 per cwt. at A and sells at B, using the same

spring balance. What must be his charge so that he may gain 20 per

cent. ?

Ans. 12. Os. 9d.

PHYSICAL INDEPENDENCE OF FORCES. 69

79. Physical Independence of Forces. The latter

part of the Second Law states that the change of motion

produced by a force is in the direction in which the force

acts.

Suppose we have a particle in motion in the direction

AB and a force acting on it in the direction AC; then

the law states that the velocity in the direction AB is

unchanged, and that the only change of velocity is in the

direction AC; so that to find the real velocity of the

particle at the end of a unit of time, we must compoundthe velocity in the direction AB with the velocity generated

in that unit of time by the force in the direction AC.

The same reasoning would hold if we had a second force

acting on the particle in some other direction, and so for

any system of forces. Hence if a set of forces act on a

particle at rest, or in motion, their combined effect is

found by considering the effect of each force on the particle

just as if the other forces did not exist and as if the particle

were at rest, and then compounding these effects. This

principle is often referred to as that of the Physical Inde-

pendence of Forces.

As an illustration of this principle consider the motion

of a ball allowed to fall from the hand of a passenger in a

train which is travelling rapidly. It will be found to hit

the floor of the carriage at exactly the same spot as it

would have done if the carriage had been at rest. This

shews that the ball must have continued to move forward

with the same velocity that the train had or, in other

words, the weight of the body only altered the motion in

the vertical direction, and had no influence on the horizontal

velocity of the particle.

70 PARALLELOGRAM OF FORCES.

80. Parallelogram of Forces. We have shewn

in Art. 28 that if a particle of mass m have accelerations f, ,

fz represented in magnitude and direction by lines AB, AC,then its resultant acceleration fa

is represented in magni-tude and direction by AD, the diagonal of the parallelogramof which AB, AC are adjacent sides.

Since the particle has an acceleration f, in the direction

AB there must be a force P, (= m/j) in that direction, and

similarly a force P2 (= w/2)

in the direction AC. Let

AB1} AC, represent these forces in magnitude and direc-

tion. Complete the parallelogram AB,D,C,. Then since

the forces in the directions AB,, AC, are proportional to

the accelerations in those directions,

/. AB, : AB :: B,D, : BD.

Hence by simple geometry we have A, D, D, in a

straight line, and

/. AD, : AD :: AB, : AB.

.'. AD1 represents the force which produces the acceleration

represented by AD, and hence is the force which is equiva-

lent to the forces represented by AB,, AC,.Hence we infer the truth of the Parallelogram of

Forces which may be enunciated as follows;

If a particle be acted on by two forces represented in

magnitude and direction by the two sides of a parallelogram

LAW III. 71

drawn from a point, they are equivalent to a force repre-

sented in magnitude and direction by the diagonal of the

parallelogram passing through the point.

Cor. If in Arts. 17 23 which are founded on the

Parallelogram of Velocities we substitute the word "force"

for"velocity

"they will still be true.

81. The student will notice that since a force has

direction as well as magnitude it, like a velocity or an

acceleration, is a vector quantity. All vector quantities will

be found to be compounded according to the parallelogram

law, whilst scalar quantities are compounded by simple

addition.

82. Law III. To every action there is an equal and

opposite reaction.

Every exertion of force consists of a mutual action

between two bodies. This mutual action is called the

stress between the two bodies, so that the Action and

Reaction of ^Newton together form the Stress.

If a book rest on a table, the book presses the table

with a force equal and opposite to that which the table

exerts on the book.

If a man raise a weight by means of a string tied to

it, the string exerts on the man's hand exactly the same

force that it exerts on the weight, but in the opposite

direction.

The attraction of the earth on a body is its weight,

and the body attracts the earth with a force equal and

opposite to its own weight.

When a horse drags a canal-boat by means of a rope,

the rope drags the horse back with a force equal to that

with which it drags the boat forward.


83. Motion of two particles connected by a

string passing over a small smooth pulley.

Two particles of masses m1 ,m

2are connected by a light

inextensible string which passes over a small

smooth pulley. Ifm 1be> m

2 , find the resultingmotion of the system, and the tension of the

string.

Let the tension of the string be T poundals;the pulley being smooth, this will be the same

throughout the string.

Since the string is inextensible, the velocity

of m3 upwards must, throughout the motion, 2

be the same as that of mldownwards.

Hence their accelerations [rates of changeof velocity] must be the same in magnitude.Let the magnitude of the common acceleration be /.

Now the force on mldownwards is m^g T poundals.

Hence mtg T = m

tf. (i).

So the force on ra2 upwards is T m^g poundals ;

/. T-mtg = mj. (ii).

Adding (i) and (ii) we have /= ! ?q. giving- them

i + mz

common acceleration.

Also from (ii) T=m^(f+g)= - f1*

g poundals.ml + 7/i

2

Since the acceleration is known and constant, the

equations of Art. 41 give the space moved through andthe velocity acquired in any given time.

Cor. 1. If the tension of the string is equal to the

weight ofM Ibs., or to Mg poundals, we have

ATWOOD S MACHINE. 73

211Hence ^ = h ,

and the tension of the string is/I// ofYt KYI *~*Ml lll

l//<-

2

therefore equal to the weight of a mass, which is the

harmonic mean between the two masses.

Cor. 2. Since the harmonic mean between two

quantities is always less than the arithmetic mean, it

follows that the tension of the string is less than the

semi-sum of the weights of the masses. Hence the

pressure on the axle of the pulley, which is equal to twice

the tension of the string, is less than the sum of the

weights of the masses.

84. Atwood's machine. This machine in its sim-

plest form consists of a vertical pillar

AB, of about 8 feet in height, firmly

clamped to the ground, and carryingat its top a light pulley which will

move very freely. This pillar is gradu-ated and carries two platforms D, F,

arid a ring E, all of which can be

affixed by screws at any height de-

sired. The platform D can also be

instantaneously dropped. Over the

pulley passes a fine cord supportingat its ends two long thin equal weights,one of which, P, can freely pass throughthe ring E. Another small weight Qis provided, which can be laid uponthe weight P, but which cannot pass

through the ring E.

The weight Q is laid upon P, the platform D is

dropped, and motion ensues;the weight Q is left behind

74 ATWOOD'S MACHINE.

as the weight P passes through the ring; the weight Pthen traverses the distance EF with constant velocity, and

the time T which it takes to describe this distance is

carefully measured.

By the last article, the acceleration of the system as

the weight falls from D to E is 'ff or ~

Denote this by/and let DE = h.

Then the velocity v on arriving at E is given by

After passing E, the distance EF is described with

constant velocity v.

Hence, if EF = h1}we have

T = ^ =

Since all the quantities involved are known, this relation

gives us the value of g.

By giving different values to P, Q, h, h1}we can in

this manner verify all the fundamental laws of motion.

In practice, the value of g cannot by this method be

found to any great degree of accuracy ;the chief causes of

discrepancy being the mass of the pulley, which cannot be

neglected, the friction of the pivot on which the wheel

turns, and the resistance of the air.

The effect of the pulley may be allowed for by taking

the acceleration equal to ^ ^ ^g, where N is aZtlr + (jj + ./V

quantity to be determined by comparing the results of

successive experiments.

EXAMPLES. 75

The friction of the pivot may be minimised if its ends

do not rest on fixed supports but on the circumferences of

four light wheels, called friction wheels, two on each side,

which turn very freely.

There are other pieces of apparatus for securing the

accuracy of the experiment as far as possible, e.g. for

instantaneously withdrawing the platform D at the

required moment, and a clock for beating seconds accu-

rately.

EXAMPLES.

1. A particle slides down a rough inclined plane inclined to the horizon

at an angle a; if fj.be the coefficient offriction, determine the motion.

Let m be the mass of the particle so that its weight is mg poundals ;

R the normal reaction, and JJ.R the force of friction.

Then the total force perpendicular to the plane is

(R mg cos a) poundals.

Also the total force down the plane is (mg sin a - pR) poundals.Now perpendicular to the plane there cannot be any motion, and

hence there is no change of motion.

Hence the acceleration, and therefore the force, in that direction is

zero.

.'. R-mg cosa = (i).

76 EXAMPLES.

Also the acceleration down the plane

moving force mq sin a - u,R= T = =g (sm a- a cos a), by (i).mass moved mHence the velocity of the particle after it has moved from rest over a

length I of the plane is by Art. 41 equal to

(sin a.- n cos a).

Similarly, if the particle were projected up the plane, we have to

change the sign of p, and its acceleration in a direction opposite to that

of its motion is #(sma + ;ucosa).

2. Two equally rough inclined planes, of equal height, whose inclinations

to the horizon are als o2 , are placed back to back; two masses m15 m2 , are

placed on their inclined faces and connected by a light inextensible string

passing over a smooth pulley at the common vertex of the two planes; i/mjdescend, Jind the resulting motion,

Let T poundals be the tension of the string and /j. the coefficient of

friction.

Then, as in the last example, the force on m^ along the plane is

ma <7 (sin Oj-

/it cos ax)- T ; and that on m2 is T - m$ (sin a2 + /x cos a2).

Hence if/be the magnitude of the common acceleration, we have

m1g(sina1 -fj.cosa1)-Tm1f. ....................... (i),

T-m^g (sino2 + /icosa2)= m2/ ..................... (ii),

/. by addition

f9 D'h. 8U1 ai~ m2 s*n a2

~ M (mi cos ai +m2 cos Os)]/!771! +m3l'

By substituting this value of/in (i) we have the value of T.

3. A train of mass 150 tons, moving at the rate of 30 miles per hour,

comes to the foot of an incline of 1 in 100, and the engine exerts a constant

force on the train until it arrives at the top of the incline, which is one mile

in length, when the steam is shut off, and the train runs half a mile before

coming to rest. Assuming the total resistance due to friction and the

resistance of the air to be 8 Ibs. per ton, find the force exerted by the

engine.

When we say that the resistance is 8 Ibs. per ton we mean that for

every ton in the mass of the train the resistance is equal to the weight of

8 pounds.

/. total resistance = weight of 1200 Ibs. = 1200</ poundals.

The resolved part of the weight down the plane

-poundals= 3360# poundals.

EXAMPLES. 77

Let the force exerted by the engine be equal to the weight of x tons or

to 2240 x x x g poundals.

.-. total acceleration up the plane

= (2240x- 3360 - 1200) g+ 150 . 2240

/ x 19 \n I I r i ao.v\- g\i5o uw)~J( J) '

Also the velocity at the foot of the incline is 44 feet per second.

Let v be the velocity on arriving at the top of the incline. Then byArt. 41

t>a=442 + 2/. 1760.3 (i).

On the level at the top the acceleration is -^-

- or -j . Also

150 . 2240 280after running half a mile the velocity is zero.

9

Equating the two values of v in (i), (ii)we get a; = l|.

Hence the tractive force is equal to the weight of 1^| tons.

4. A body whose mass is n pounds is placed on a horizontal plane which

is in motion in a vertical direction; if the pressure of the body on the plane

be equal to the weight of m pounds, find the acceleration with which the

plane is moving.

Since the pressure of the body on the plane is equal to the weight of

m Ibs. or to ing poundals, therefore by the third law of motion the pressure

of the plane on the body is mg poundals in the opposite direction.

The only other force acting on the body is its weight, which is ng

poundals downwards.

/. the resultant force on the body is (m -n) g poundals vertically

upwards.

Hence the acceleration of the body

moving force m - n

mass moved -g vertically upwards.

Cor. If m be >n, i.e. if the pressure of the body on the plane be

greater than its weight, this acceleration is positive ; hence the acceleration

of the plane is vertically upwards, and it is either moving upwards with

an increasing velocity or downwards with a decreasing velocity.

If m be <7i, this acceleration is negative; hence the acceleration of

the plane is downwards, and it is either moving upwards with a decreasing

velocity or downwards with an increasing velocity.

78 EXAMPLES.

5. A mass of 9 Ibs. descending vertically drags up a mass of 6 Ibs.

by means of a string passing over a smooth pulley ;find the acceleration

of the system and the tension of the string.

Am. f ; weight of 1\ Ibs.5

6. Two masses each equal to m are connected by a string passing

over a smooth pulley ;what mass must be taken from one and added to

the other so that the system may describe 200 feet in 5 seconds?

mAn*.

2-

7. Two equal masses, of 3 Ibs. each, are connected by a light string

hanging over a smooth peg ;if a third mass of 3 Ibs. be laid on one of

them, by how much is the pressure on the peg increased?

Ans. By the weight of 2 Ibs.

8. A mass of 4 ozs. is attached by a string passing over a smooth

pulley to a larger mass; find the magnitude of the latter so that, if after

the motion has continued 3 seconds the string be cut, the former will

ascend -^ ft. before descending.

Ans. 5 ozs.

9. Two scalepans of mass 3 Ibs. each are connected by a string

passing over a smooth pulley ;shew how to divide a mass of 12 Ibs.

between the two scalepans so that the heavier may descend a distance of

50 feet in the first five seconds.

Ans. The mass must be divided in the ratio 19 : 13.

10. Two strings pass over a smooth pulley; on one side they are

attached to masses of 3 and 4 Ibs. respectively, and on the other to one of

5 Ibs.;find the tensions of the strings and the acceleration of the system.

Ans. Weights of 2^ and 3 Ibs. respectively ; |.

11. A string hung over a pulley has at one end a mass of 10 Ibs.

and at the other end masses of 8 and 4 Ibs. respectively ;after being in

motion for 5 seconds the four pound mass is taken off; find how much

further the masses go before they first come to rest.

Ans. 29ft. 9 ins. nearly.

12. A mass m pulls a mass m' up an inclined plane, inclination a,

by means of a string passing over a pulley at the top of the plane ;shew

that the acceleration is

m-m' sin a

m+m' g '

EXAMPLES. 79

13. A mass of 12 Ibs. drags a mass of 16 Ibs. up an inclined plane,

whose inclination is 30, being attached to it by means of a string passing

over the top of the plane ;find the distance described in 5 seconds and

the tension of the string.

. 57| feet; weight of lOf Ibs.

14. A mass of 6 ounces slides down a smooth inclined plane, whose

height is half its length, and draws another mass from rest over a distance

of three feet in five seconds along a horizontal table which is level with

the top of the plane over which the string passes; find the mass on the

table.

Ans. 24 Ibs. 10 ozs.

15. A mass of 5 Ibs. on a rough horizontal table is connected by

a string with a mass of 8 Ibs. which hangs over the edge of the table;

find the coefficient of friction in order that the heavier mass may move

vertically with half the acceleration it would have if it fell freely.

Ans. /*= '3.

16. A mass Q on a rough horizontal plane, coefficient of friction >J3,

is connected by a string with a mass 3Q which hangs over the edge of the

table; if motion ensue and the string break after four seconds, find

how far the new position of equilibrium of Q will be from its initial

position.

Ans. 96 feet.

17. Sixteen balls of equal mass are strung like beads on a string ;

some are placed on an inclined plane of inclination sin" 1J, and the rest

hang over the top of the plane ; how have the balls been arranged if the

acceleration at first is ^ ?m

Ans. 10 balls must hang vertically.

18. A mass P is drawn up a smooth plane inclined at an angle of

30 to the horizon by means of a mass Q which descends vertically, the

masses being joined by a string passing over a smooth pulley at the top

of the plane ;if the acceleration of the system be one-fourth that of a

freely falling body find the ratio of Q to P.

Ans. They are equal.

19. A man whose weight is 8 stone stands on a lift which moveswith a uniform acceleration of 12 ft. -sec. units; find the pressure on the

floor when the lift is (1) ascending, (2) descending.

Ans. (I) 154 Ibs. weight ;

'

(2) 70 Ibs. weight.

80 EXAMPLES.

20. A balloon ascends with a uniformly accelerated velocity, so that

a mass of one cwt. produces on the floor of the balloon the same pressure

which 116 Ibs. would produce on the earth's surface; find the height

which the balloon will have attained in one minute from the time of

starting.

Ans. 2057fft.

21. A bucket containing one cwt. of coal is drawn up the shaft of a

coal-pit and the pressure of the coal on the bottom of the bucket is equal

to the weight of 126 Ibs. Find the acceleration of the bucket.

Ans. | ft.-sec. units.o

22. A bullet with an initial velocity of 1500 feet per second strikes a

target at 1200 yards distance with a velocity of 900 feet per second; the

range of the bullet being supposed horizontal, compare the mean resistance

of the air with the weight of the bullet.

Ans. They are as 25 : 4.

23. A train of mass 200 tons is running at the rate of 40 miles per

hour down an incline of 1 in 120;find the resistance necessary to stop it

in half a mile.

Ans. The weight of 5ff tons.

24. A train runs from rest for one mile down a plane whose descent

is one foot vertically for each 100 feet of its length ;if the resistances be

equal to 8 Ibs. per ton how far will the train be carried along the horizontal

level at the foot of the incline?

Ans. 1 mile 1408 yards.

25. P hangs vertically and is 9 Ibs. ; Q is a mass of 6 Ibs. on a plane

whose inclination to the horizon is 30 ; shew that P will drag Q up the

whole length of the plane in half the time that Q hanging vertically would

take to draw P up the plane.

26. A mass m is drawn up an inclined plane of height h and length I

by means of a string passing over the vertex of the plane, from the other

end of which hangs a mass m'. Shew that in order that m may just

reach the top of the plane, m' must be detached after m has moved

through a distance

m + m' hi

EXAMPLES. 81

27. Two masses are connected by a string passing over a small

pulley; shew that if the sum of the masses be constant, the tension of

the string is greater, the less the acceleration.

28. A mass A draws a mass B up an inclined plane by means of a

string passing over the vertex;find the inclination of the plane so that A

may draw B up a given vertical height in the least time.

Ans. The inclination must be sin"1.

Aft

29. A plane of height h and inclination a to the horizon has cut in

it a groove inclined at an angle /3 to the line of greatest slope ;find the

time a particle would take to describe the groove starting from rest at

the top.

/2/tV7Ans, A / cosec a . sec B.> 9

30. A train of mass 50 tons is moving on a level at the rate of

30 miles per hour when the steam is shut off, and the brake being

applied to the brake-van the train is stopped in a quarter of a mile. Find

the mass of the brake-van, taking the coefficient of friction between the

wheels and rails to be one-sixth and supposing the unlocked wheels to roll

without sliding.

Ans. 6 tons.

85. Impulse. Def. The impulse of a force in a

given time is proportional to the product of the force (if

constant, and the mean value of the force if variable} and

the time during which it acts.

If the unit of impulse be the impulse of a unit force

acting for the unit of time, then the impulse of a force Facting for a time t is F . t.

The impulse of a force is also equal to the momentum

generated by the force in the given time. For suppose a

particle, of mass m, moving initially with velocity u is

acted on by a constant force F for time t. If f be the

resulting acceleration we have F=mf.L. D. 6

82 IMPULSE.

But if v be the velocity of the particle at the end of

time t, we havev = u +ft.

Hence the impulse = Ft = mft = mv mu

.= the momentum generated in the given time.

The same result is true if the force be variable. For

suppose the time t to be divided into a very large number

(ri) of portions each equal to r, and let these portions of

time be taken so small that in each of them the force

may be considered to remain appreciably constant.

Let the values of the force be FI}F

2 , ...... Fn ,and the

resulting accelerations be/t ,/2...... fn .

Then we have

Hence the impulse = mean force x t

F +F . . +F__J.

1~r J.

z...... ~ * n /

= m(/1r+/2T+= m x sum of changes of the velocity during the

whole interval

= momentum generated in time t.

From the preceding article it follows that the second

law of motion might have been enunciated in the following

form;

The change of momentum of a particle in a

given time is equal to the impulse of the force

which produces it and is in the same direction.

This is the form of the second law usually quoted.

IMPULSIVE FORCES. 83

86. Impulsive Forces. Suppose we have a force

F acting for a time T on a body whose mass is m, and let

the velocities of the mass at the beginning and end of this

time be u and v. Then by the last article

FT m(v u).

Let now the force become bigger and bigger and the time

T smaller and smaller. Then ultimately F will be almost

infinitely big and T almost infinitely small, and yet their

product may be finite. For example F may be equal to

107

poundals, r equal tor-^ seconds, and m equal to one

pound, in which case the change of velocity produced is

the unit of velocity.

To find the whole effect of a finite force acting for a

finite time we have to find two things, (1) the change in

the velocity of the particle produced by the force duringthe time it acts, and (2) the change in the position of

the particle during this time. Now in the case of an

infinitely large force acting for an infinitely short time,

the body moves only a very short distance whilst the force

is acting on it, so that the change of position of the particle

may be neglected. Hence the total effect of such a force

is known when we know the change of momentum which

it produces.

Such a force is called an impulsive force. Hence

Def. An impulsive force is a very great force acting

for a very short time, so that the change in the position ofthe particle during the time the force acts on it may be

neglected. Its whole effect is measured by its impulse, or

the change of momentum produced.

In actual practice we never have any experience of an

G 2

84 IMPACT OF TWO BODIES.

infinitely great force acting for an infinitely short time.

Approximate examples are, however, the blow of a

hammer, the collision of two billiard balls.

The above will be true even if the force is not uniform.

In this case we must define F as the mean force. In the

ordinary case of the collision of two billiard balls the force

generally varies very considerably.

We may notice that ifwe have both finite and impulsive

forces acting on a body at the same time the former maybe neglected in comparison with the latter

;for the effect

of a finite force during an infinitely short time is infinitely

small;a finite force necessarily requires a finite time to

produce a finite effect. Hence if we are considering the

effect of a collision between two balls which are in motion

in the air, we can neglect the effect of gravity during the

time that the collision lasts.

Ex. 1. A body, whose mass is 9 Ibs., is acted on by a force which

changes its velocity from 20 miles per hour to 30 miles per hour. Find

the impulse of the force.

Ans. 132 units of impulse.

Ex. 2. A mass of 2 Ibs. at rest is struck and starts off with a velocity

of 10 feet per second;

if the time during which the blow lasts be T^thof a second find the average value of the force acting on the mass.

Ans. 2000 poundals.

Ex. 3. A glass marble whose mass is one ounce falls from a height

of 25 feet and rebounds to a height of 16 feet ;find the impulse and the

average force between the marble and the floor if the time during which

they are in contact is ^th of a second.

Ans. 4| units of impulse ;45 poundals.

87. Impact of two bodies. When two masses A,

B impinge, then by the third law of motion the action of

MOTION OF A SHOT AND GUN. 85

A on B is, at each instant during which they are in contact,

equal and opposite to that of B on A.

Hence the impulse of the action of A on B is equal

and opposite to the impulse of the action of B on A.

It follows that the change in the momentum of is

equal and opposite to the change in the momentum of A,

and therefore the sum of these changes measured in the

same direction is zero.

Hence the sum of the momenta of the two masses

measured in the same direction is unaltered by their

impact.

88. Motion of a shot and gun. When a gun is

fired the powder is almost instantaneously converted into

a gas at a very high pressure which by its expansion

forces the shot out. The action of the gas is similar to

that of a compressed spring trying to recover its natural

position. The force exerted on the shot forwards is at anyinstant before the shot leaves the gun equal and opposite

to that exerted on the gun backwards, and therefore the

impulse of the force on the shot is equal and opposite to

the impulse of the force on the gun. Hence the momentum

generated in the shot is equal and opposite to that generatedin the gun, if the latter be free to move.

Ex. 1. A body, of mass 3 Ibs., moving with velocity 13 feet per

second overtakes a body, of mass 2 Ibs., moving with velocity 3 feet per

second in the same straight line and they coalesce and form one body ;

find the velocity of this single body.

Let V be the required velocity. Then since the sum of the momentaof the two bodies is unaltered by the impact, we have

(3 + 2) V= 3x 13 + 2x3 = 45 units of momentum,

/. F=9ft. per sec.

86 WORK.

Ex. 2. A body of mass 7 Ibs., moving with a velocity of 10 feet per

second overtakes a body, of mass 20 Ibs., moving with a velocity of 2 feet

per second in the same direction as the first;

if after the impact they

move forward with a common velocity, find its magnitude.

Ans. 4-,j2T ft. per sec.

Ex. 3. A body, of mass 10 Ibs., moving with velocity 4 feet per

second meets a body of mass 12 Ibs. moving in the opposite direction

with a velocity of 7 feet per second;

if they coalesce into one body, shew

that it will have a velocity of 2 feet per second in the direction in which

the larger body was originally moving.

Ex. 4. A shot of mass 1 ounce is projected with a velocity of 1000

feet per second from a gun of mass 10 Ibs.;find the velocity with which

the latter begins to recoil.

Ans. 6 ft. per sec.

89. Work. Def. A force is said to do work when it

moves its point of application.

The measure of this work is the product of the force

and the distance through which the point of application is

moved in the direction of the force.

It will be noticed that this definition says nothing about

the time occupied in changing the position of the point of

application, or of the path by which the point of applica-

tion is brought from the first into the second position.

Let a force F acting in the direction AB have its

point of application moved from A to C;

draw CD

perpendicular to AB;then the work done by the force is

proportional to the product of F and AD. When the

point D is on the side of A toward which the force acts

then the displacement AD, and the work done by the

force are said to be positive.

WORK. 87

When however the point D is on the opposite side as

c

o A F

in the second figure the work done, and the displacement,are said to be negative. This is sometimes expressed by

saying that the force has work done against it. If a manraise a body from the ground, the work done by the manis positive whilst that done by the body is negative.

If the angle CAD be a right angle, then the point Dcoincides with A, and the displacement AD, and therefore

the work done by the force, are both zero. Hence the

work done by a force when its point of application is

moved in a direction perpendicular to the direction of the

force is zero.

No work is done by the weight of a body which is

moved about on a horizontal table.

If a body be sliding down an inclined plane no work is

done by or against the pressure of the plane on the body.

90. The British absolute unit of work is the work done

by a poundal in moving its point of application through one

foot, and is called a Foot-Poundal. It is roughly equalto the work done in lifting half an ounce through one foot.

With this unit the work done by a force of F poundals in

moving its point of application through s feet is Fs units

of work.

The c. G. s. unit of work is that done by a dyne in

moving its point of application through a centimetre, and

is called an Erg. On account of the smallness of this

unit a larger unit is sometimes used and is called a Joule.

One Joule is equal to 107

ergs.

88 WORK.

One Foot-Poundal = 421390 Ergs nearly.

The practical unit used in England by engineers is

called a Foot-Pound and is the work done in raising the

weight of a pound vertically through one foot. The

weight of a pound differs at different points of the earth's

surface, so that this is not a suitable unit for theoretical

calculations.

One Foot-Pound = g . Foot-Poundals.

The corresponding unit used by French engineers is

the gramme-centimetre, which is the work done in raising

the weight of a gramme through one centimetre vertically.

91. Def. The rate of work or Power of an agent is

the amount of work that would be done by the agent in a

unit of time.

An agent is working with the British absolute unit of

power when it does a foot-poundal per second.

Similarly it is working with the c. G. s. absolute unit of

power when it does an erg per second. When it is per-

forming 1 Joule or 107

ergs per second it is said to be

working with a power of one Watt.

The unit of power used by engineers is a Horse-Power. An agent is said to be working with one Horse-

Power or H.-P. when it performs 33000 foot-pounds in a

minute, i.e. when it raises 33000 pounds vertically throughone foot per minute. This estimate of the power of a

horse was made by Watt but is rather above the capacityof ordinary horses.

The unit of power used by French engineers is called

a Force de cheval. It is equivalent to 75 kilogramme-metres per second or to about 542 foot-pounds per second.

Thus it is a little less than one horse-power.

EXAMPLES. 89

EXAMPLES.

1. Find the work done in dragging a heavy body up an inclined

plane.

Let AB be the inclined plane, J5D the perpendicular drawn from B on

the horizontal plane through A, and let the angle BAD = a.

If Hrbe the weight of the body the resolved part of the weight down

the plane is W sin a, and the resolved part perpendicular to the plane is

TFcosa.

Hence the resistance due to friction during the motion is /^Kcosa,

where /t is the coefficient of friction. Hence the total force to be overcome

is W (sin a + fj.cos a).

/. work done in dragging the body up a distance AB=W (sin a + fji cos o) . AB= W . DB + pW . AD= work done in raising the weight through a vertical distance DB, together

with the work done in dragging it a distance equal to AD along a hori-

zontal plane of the same roughness as the inclined plane.

2. Find the work done in stretching an elastic string.

Let OA ( o) be the unstretched length of the string, and let us find

the work done in stretching it from a length OB (=

b) to 0(7(=

c).

When the length is OP, the tension =X(OP -

a) = - PA.a v 'a

B

Erect at P a perpendicular PQ to represent this tension.

PQ \Then -~ -

; hence the angle PA Q is constant, and therefore QITA CL

always lies on a straight line through A . If this meet the perpendiculars

through B, C in E, F then the required work is, as in Art. 58, represented

by the area BEFC, and hence is %BC x (BE+CF) or is equal to the

Extension produced multiplied by the mean of the initial and finaltensions.

90 EXAMPLES.

3. A train of mass 100 tons is ascending uniformly an incline of 1 in

280 and the resistance due to friction etc. is equal to 16 Ibs. per ton; if the

engine be 0/200 H.-P. and be working at full power, find the rate at which

the train is going.

The resistance due to friction etc. is equal to the weight of 1600

pounds, and the resolved part of the weight of the train down the incline

is equal to the weight of -^^ of 100 tons, or to the weight of 800 Ibs. ;

hence the total force to impede the motion is equal to the weight of

2400 Ibs.

Let v be the velocity of the train in feet per second. Then the work

done by the engine is that done in dragging a force equal to the weight of

2400 Ibs. through v feet per second and is equivalent to 2400i; foot-

pounds per second.

But the total work which the engine can do is ^/p- r 110,000

foot-pounds per second.

Hence 2400v = 110000,

1100V=-2T>

and hence the velocity of the train is 31 miles per hour.

4. A train of mass 80 tons is, at a certain instant, moving at the rate

of 30 miles per hour up an incline of 1 in 100, the resistances due to

friction etc. being equal to 14 Ibs. per ton; if the engine be 0/250 horse-

power and be working at full power, find the acceleration of the train at

that instant.

The train is moving at the rate of 44 feet per second, and the engine

is doing 250 x 550 foot-pounds per second, and is therefore exerting a force

equal to the weight of - T-T or 3125 pounds.

Now the force down the plane due to the weight of the train and the

friction= 80 [2240-7-100 + 14] pounds weight

= 2912 pounds weight.

.-.the force to accelerate the train= 213 pounds weight

= 213x32poundals.

213 x 32 213

EXAMPLES. 91

5. Find the H.-P. of an engine which will travel at the rate of

25 miles per hour up an incline of 1 in 100, the mass of the engine and

load being 10 tons and the resistances being 10 Ibs. per ton.

Am, 21$.

6. A train of mass 200 tons, including the engine, is drawn up an

incline of 3 in 500 at the rate of 40 miles per hour by an engine of 600

H.-P. : find the resistance per ton due to friction etc.

Ana. 14-685 Ibs. weight.

7. A weight of 10 tons is dragged in half-an-hour a length of 330 feet

up a rough plane inclined at an angle of 30 to the horizon ; the coefficient

of friction being -T^ find the work expended, and the H.-P. of an engineV

by which it will be done.

Ans. 7,392,000 ft. -Ibs.;7'46 H.-P.

8. What is the horse-power of an engine which keeps a train of

100 tons going at the rate of 40 miles per hour against a resistance of

2000 Ibs.?

Ans. 213i H.-P.

9. The least H.-P. of an engine which is taking a train of 120 tons

up an incline of 1 in 224 at the rate of 30 miles per hour is 336, supposing

a resistance of 25 Ibs. per ton on the level at this speed.

10. Shew that the work done in raising a given mass M from one

position to another is Mgh, where h is the distance through which the

centre of inertia of the mass has been raised vertically.

11. In how many hours would an engine of 18 H.-P. empty a

vertical shaft full of water, if the diameter of the shaft be 9 feet, the depth

420 feet, and the mass of a cubic foot of water 62 Ibs.?

Ans. 9f| hours.

12. An elastic string is gradually stretched by an increasing force;

when the string has been lengthened 2 inches the tension is found to be

equal to the weight of 10 Ibs. What is the work that has been done in

stretching the string?

Ans. %g foot-poundals.

92 ENERGY.

92. Energy. Def. The Energy of a body is its

capacity for doing work and is of two kinds, Kinetic and

Potential.

The Kinetic Energy of a body is the energy which it

possesses by virtue of its motion and is measured by the

amount of work that the body can perform against the im-

pressed forces before its velocity is destroyed.

A falling body, a swinging pendulum, and a cannon

ball in motion all possess kinetic energy.

Consider the case of a particle, of mass m, movingwith velocity u, and let us find the work done by it before

it comes to rest.

Suppose it brought to rest by a constant force F

resisting its motion, which produces in it an acceleration

/ given by F= mf.

Let x be the space described by the body before it

comes to rest so that

Hence the kinetic energy of the body

= work done by it before it comes to rest

= Fx = mfx = ^mu2

.

Hence the kinetic energy is equal to the product of the

mass of the body and one half the square of its velocity.

The definition given above therefore coincides with

that given in Art. 68.

93. The Potential Energy of a body is the work it

can do by means of its position in passing from its present

configuration to some standard configuration (usually called

its zero position).

CONSERVATION OF ENERGY. 93

A bent spring has potential energy, viz. the work it

can do in recovering its position. A body raised to a

height above the ground has potential energy, viz. the

work its weight can do as it falls to the earth's surface.

The following example is important :

94. A particle of mass m falls from rest at a height h

above the ground ; to shew that the sum of its potential and

kinetic energies is constant throughout the motion.

Let H be the point from which the particle starts and

the point where it reaches the ground.

Let v be its velocity when it has fallen through a

distance HP (= x), so that v* = Igx.

:. kinetic energy there = ^mv2 = mgx.

Also its potential energy at P= the work its weight can do as it falls from P to

= mg . PO = mg (h x).

:. the sum of its kinetic and potential energies at P= mgh.

But its potential energy when at H is mgh and its kinetic

energy there is zero.

Hence the sum of the potential and kinetic energies is

the same at P as at H; and since P is any point it

follows that the sum of these two quantities is the same

throughout the motion.

95. The above is an extremely simple example of the

principle of the Conservation of Energy, which may be

stated as follows :

If a body or system of bodies be in motion under a

conservative system of forces, the sum of its kinetic and

potential energies is constant.

94 CONSERVATION OF ENERGY.

A conservative system of forces is one which depends on

the position or configuration only of the system of bodies,

and not on the velocity or direction of motion of the bodies.

Thus from a conservative system are excluded forces

of the nature of friction, or forces such as the resistance of

the air which varies as some power of the velocity of the

body. Friction is excluded because if the direction of

motion of the body be reversed, the direction of the friction

is reversed also.

Referring to the case of a particle sliding down a

rough plane of length I (Art. 84, Ex. 1) we see that the

kinetic energy of the particle on reaching the ground is

\m [*2gl (sin a/j,

cos a)], or mgl sin a mglfjb cos a. Also

the potential energy there is zero, so that the sum of the

kinetic and potential energies at the foot of the plane is

mgl sin a. pmgl cos a.

But the potential energy of the particle when at the topof the plane is mg . I sin a, so that the total loss of visible

mechanical energy of the particle in sliding from the

top to the bottom of the inclined plane is pmgl cos or.

This energy has been transformed and appears chiefly in

the form of heat, partly in the moving body, and partly in

the plane.

* 96. Theorem. To shew that the change of kinetic

energy per unit of space is equal to the acting force.

If a force F acting on a particle of mass m change the

velocity from u to v in time t whilst the particle moves

through a space s, we have v* u* = 2fs, where f is the

acceleration produced.

GRAPHIC METHOD. 95

This equation proves the proposition when the force is

constant.

When the force is variable, the same proof will hold

if we take t so small that the force F does not sensibly

alter during that interval.

Cor. It follows from equation (1) that the change in

the kinetic energy of a particle is equal to the work done

on it.

Ex. 1. Find the kinetic energy in ergs of a cannon ball of 10,000

grammes discharged with a velocity of 5000 centimetres per second.

Ana. 1-25 x 1011ergs.

Ex. 2. A cannon ball, of mass 5000 grammes, is discharged with a

velocity of 500 metres per second. Find its kinetic energy in ergs, and,

if the cannon be free to move and have a mass of 100 kilogrammes, find

the energy of the recoil.

Am. 6-25 x 1012ergs ;

3-125 x 1011ergs.

Ex. 3. A bullet of mass 50 grammes is fired into a target with a

velocity of 500 metres per second. The target is of mass one kilogrammeand is free to move ;

find the loss of energy by the impact in kilogram-

metres.

Ans. 5952/r kilogrammetres.

Ex. 4. Find the H.-P. of a steam gun which projects 100 four-poundshots in a minute with a velocity of 1200 feet per second.

Ans. 272T8T .

97. Graphic Method. Force-Space Curve.Since the kinetic energy acquired by a particle in a given

time is equal to the work done on the particle in that time,

it follows that if in the figure of Art. 58 the abscissae

represent the distances described by a particle, and the

ordinates represent the corresponding forces acting on the

particle, then the kinetic energy generated is represented

by the number of units of area in OACB.

96 MOTION OF THE CENTRE OF INERTIA

Ex. 1. If a particle be moving in a straight line under a force to a

fixed point in the straight line, which varies as the distance of the

particle from the fixed point, find the kinetic energy of the particle

when a given distance has been described from rest.

Ex. 2. If in the last example, the force vary as the square of the

distance of the particle from the fixed point, find the kinetic energy

acquired.

Motion of the centre of inertia of a system ofparticles.

98. Theorem. If the velocities at any instant of anynumber ofmasses mv m2 ,

m8 parallel to any linefixed in

space be u,, ua ,u

3then the velocity parallel to that

line of the centre of inertia of these masses at that instant is

m,u t+ m2

u2 + . . .

m, + m2 + . . .

At the instant under consideration, let xv #2 ,x

3

be the distances of the given masses measured along this

fixed line from a fixed point in it and let x be the distance

of their centre of inertia.

Then (as is shewn in any treatise on Statics)

X =

Let#j',

#2

'

...... be the corresponding distances of these

masses at the end of a small time t and x' the correspondingdistance of their centre of inertia. Then we have

Also

OF A SYSTEM OF PARTICLES. 97

But if u be the velocity of the centre of inertia parallel

to the fixed line we have x = x + ut.

u =x' x m.u. + m.u. +

- = -.

t m, +mz + ......

Hence the velocity of the centre of inertia of a system of

particles in any given direction is equal to the sum of the

momenta of the particles in that direction divided by the

sum of the masses of the particles.

Cor. If a system of particles be in motion in a plane,

and their velocities and directions of motion be known,

we can by resolving these velocities parallel to two fixed

lines and applying the preceding proposition find the

motion of their centre of inertia.

So if we have a system of particles in motion in space

we must resolve the velocities along three fixed lines not

lying in one plane and proceed as before.

99. Theorem. If the accelerations at any instant of

any number of masses mi;m

2 ,m

3...... parallel to any line

fixed in space be fx ,

fa ,

fs

...... then the acceleration of the

centre of inertia of these masses parallel to this line is

The proof of this proposition is similar to that of the

last article. We have only to change #1?

ult #/, ?*/ into

ui> /i> ui> fi> and make similar changes for the other

particles.

100. Theorem. The motion of the centre of inertia

of a system of particles is unaffected by any mutual action

between the particles of the system.

L. D. 7

98 MOTION OF THE CENTRE OF INERTIA

For consider any two particles mt ,m

2of the system

and the stress between them.

[This stress may be of the nature of an impact, as

when two billiard balls impinge, or of the nature of an

attraction as when the earth and a falling body attract

one another.]

Then, by the 3rd law, the action on mxat any instant

is equal and opposite to that on m2 ;

hence the impulseof the action on m

lis equal and opposite to the impulse

of the action on m2

.

It follows that the change in the momentum of mldue

to this mutual action

= [change in the momentum of m2due to this action].

Hence the sum of the changes in their momenta due

to this action is zero, or the sum of the momenta of mv mt

is not altered by their mutual action.

Hence the sum of the momenta of the two masses

resolved along any fixed line is unaffected by the stress

between them.

The same result holds for every such pair of particles.

But the velocity of the centre of inertia in any direction

momentum of the particles in that direction

sum of their masses

Hence the velocity of the centre of inertia is unaltered

by the stresses of the system.

Cor. It follows that the first law of motion might be

enunciated as follows;The centre of inertia of any system

of particles will continue at rest, or in uniform motion in

a straight line, except it be compelled by external forces

to change that state of rest or uniform motion.

OF A SYSTEM OF PARTICLES. 99

101. Theorem. The motion of the centre of inertia of

any system of particles is the same as if all the masses of

the system were collected at the centre of inertia, and all

the external forces of the system applied there parallel to

their original directions.

Let the masses of the different particles of the systembe m

1 ,m

3...... and let the components of the external

forces acting on them parallel to any line fixed in space

Now the rate of change of the momentum of each

particle is equal to the force acting on it parallel to

the fixed direction.

/. the rate of change of the sum of the momenta of

the particles is equal to the sum of the forces acting on

the particles parallel to the fixed direction.

But these latter forces consist of the external forces of

the system together with the mutual actions of the system,and these latter are in equilibrium inter se.

Hence

where /1} /2...... are the accelerations of the particles in

the given direction.

Let f be the acceleration of the centre of inertia in

this direction.

and this latter acceleration is the same as would be

produced by a force equal to Pl+ P

2+ ...... acting on a

mass equal to the sum of the masses.

The same would be true for any other direction, and

100 CONSERVATION OF MOMENTUM.

hence the theorem follows by compounding the accelerations

and the forces.

Cor. I. The equation (i) of the preceding article

states that the rate of change of the momentum of any

system in a given direction is equal to the components of

the external forces in that direction. Hence similarly as

in Art. 85 it follows that

The change of momentum of any system in any given

direction daring any interval of time is measured by the

sum of the impulses of the external forces in the given

direction during that interval.

Hence, also, if the sum of the external forces acting on

any system resolved in any given direction always vanish,

the total momentum of the system in that direction remains

the same throughout the motion.

This latter theorem is often referred to as the principle

of the Conservation of Momentum.

Cor. II. The motion of the centre of inertia of a

heavy chain, falling freely, is the same as that of a

particle falling freely.

EXAMPLES.

1. Two masses mlt WJ2 are connected by a light string as in Art. 83;

to find the acceleration of the centre of inertia of the system.

The acceleration of the mass JH, is -- g vertically downwards, andmj + m.2

that of m2 is the same in the opposite direction.

7?i, f, +m9 f9 I w, - m2\-

.: acceleration of the centre of inertia = ^~-~ - = l+2J '

2. Two masses move at a uniform rate along two straight lines

which meet and are inclined at a given angle ; shew that their centre of

inertia describes a straight line with uniform velocity.

EXAMPLES. 101

3. Two masses m,, m., rest on a rough double inclined plane, being

connected by a light string passing over a small smooth pulley at the

vertex of the inclined plane, the inclination of whose faces to the horizon

are respectively 60 and 30. If the angle of friction be 30, and motion

be allowed to ensue, find the ratio of m^ to m.2 when the vertical accelera-

tion of their centre of inertia is one quarter the acceleration of a freely

falling particle.

An*. M.J : ).-;, :: 3+2^/3 : 1.

4. Of three equal particles which start from the highest point of a

vertical circle, one drops down a vertical diameter, and the others slide

down chords of 60 and 120 on the same side of the diameter ;shew that

the acceleration of the centre of inertia is J/19/7 down a chord of

5. Two masses are connected by a string which passes over the topof two inclined planes of equal height placed back to back. Shew that

the path of their centre of inertia is that line which joins them when

they are in such a position that the parts of the string on the two planesare to one another as the masses at their extremities.

^ MISCELLANEOUS EXAMPLES.

1. Three inches of rain fall in a certain district in 12 hours. Assumingthat the drops fall from a height of a quarter of a mile, find the pressure

on the ground per square mile of the district due to the rain during the

storm, the mass of a cubic foot of water being 1000 ounces.

The amount of rain that falls on a square foot during the storm is \

of a cubic foot, and its mass is 250 ounces.

250 1 5.-. the mass that falls per second is ^ x ^^M or Ibs.

The velocity of each raindrop on touching the ground is

N/2X0X 440 x 3 or 16 v/330 ft. per second.

.-. the momentum that is destroyed per second is

,, 5

lb v 3^0, or p units of momentum.ob4

But the number of units of momentum destroyed per second is equal

to the number of poundals in the acting force.

102 MISCELLANEOUS EXAMPLES.

5 v'330.-. pressure on the ground per square foot =

64~ Poundals<

.-. pressure per square mile= weight of 9 x 4840 x 640 x^= weight of 41 tons approximately.

2. The scale-pans of a balance are each of mass M, and in them are

placed masses M1and M2 ; to shew that the pressures on the pans during

the motion are2M

1 (M + M2 )2M2 (M M

1 )

Mj +M2 + 2M' M

x +M2+ 2M'

respectively.

Let /be the common acceleration of the system and suppose M^M^.Then as in Art. 83 we have

/= (*/!- Jf8) gl(1M+Ml + Jfa ).

Let P be the pressure between J/j and the scale-pan on which it rests ;

then the force on the mass Mlt considered as a separate body, is M^g - P.

Also its acceleration is /.

Hence M^g -P J/j /,

3. A mass P after falling freely through a distance of &feet begins to

raise a mass Q, greater than itself, being connected with

it by means of a fine string passing over a smooth pulley.

Find the resulting motion.

Let v be the velocity of the mass P just before the

string becomes tight so that v= *J2ga.

When the string becomes tight a jerk is immediately

felt on P and similarly on Q. This jerk, consisting of a

very large force acting for an exceedingly short time, is

of the nature of an impulsive force, and its effect is

measured by the change of momentum produced. Im-

mediately after the string becomes tight, the masses are

moving with a common velocity, F.

Then the impulse of the force is measured by the

change in the momentum of P, and also by the change in that of

and these are respectively P (v- V) and QV.

Hence p(v-V) = QV,

p_ PP+Q '

r>n

and the impulse of the jerk = Q V P+Q

EXAMPLES ON CHAFfER II. 103

When the string Becomes tight the acceleration of Q is by Art. 83,

Pp ff in a downward direction, and hence Q ascends a distance x given

(P\ 3 Q-P P2

p vJ

= 2 =0a;, so that x= a _ ;,,and then turns back

and moves in the opposite direction.

4. Two balls A, B of masses mlt m2 are connected by an inextensible

string and placed on a smooth table; one of them, B, is set in motion with a

given velocity in a given direction, to find the motion of each immediately

after the string has become tight.

Immediately before the string becomes tight let the velocity of B be u

at an angle a with the string. Now the action of the string during the

impact is along AB only, and hence the velocity u sin a of B perpendicular

to AB is unaltered. Also since the string is inextensible the velocities of

A, B along the line AB immediately after the string has become tight

must be the same, V. Then immediately after the string is tight, the

momentum of the system along AB is (m 1 +m2 ) V, and this must, as in

Art. 86, be equal to m.2 u cos a.

Hence the instant after the string becomes tight the velocities of Balong and perpendicular to AB are mzu cos a/(% + r 2)

and u sin a.

EXAMPLES. CHAPTER II.

1. Two scale-pans, each of mass 2 ounces, are suspended

by a weightless string over a smooth pulley; a mass of 10

ounces is placed in the one, and 4 ounces in the other. Find

the tension of the string and the pressures on the scale-pans.

2. A shot of mass m is fired from a gun of mass M with

velocity u relative to the gun ;shew that the actual velocities

Mu , muof the shot and gun are -=7 and ^- respectively.M+m M+m

3. If a shot be fired from a gun, shew that, neglecting the

mass of the powder, the work done on the shot and gun

respectively are inversely proportional to their masses.

104 EXAMPLES ON CHAPTER II.

4. A shot, whose mass is 800 pounds, is discharged from

an 81-ton gun with a velocity of 1400 feet per second; find

the steady pressure which acting on the gun would stop it

after a recoil of five feet.

5. A gun is mounted on a gun-carriage movable on a

smooth horizontal plane and the gun is elevated at an angle a

to the horizon;

a shot is fired and leaves the gun in a

direction inclined at an angle 6 to the horizon;

if the mass of

the gun and its carriage be n times that of the shot, shew that

tan ( 1 H ) tan a.

6. Find the energy per second of a waterfall 30 yards

high and a quarter of a mile broad, where the water is 20 feet

deep and has a velocity of7-J

miles per hour when it arrives

at the fall. The mass of the water is 1024 ounces per cubic

foot.

7. A steam hammer of mass 20 tons falls vertically

through 5 feet, being pressed downwards by steam pressure

equal to the weight of 30 tons;what velocity will it acquire,

and how many foot-pounds of work will it do before comingto rest?

8. Find in horse-power the rate at which a locomotive

must be able to work in order to get up a velocity of 20 miles

per hour on a level line in a train of 60 tons in 3 minutes

after starting, the resistance to motion being taken at 10 pounds

per ton.

9. Shew that a train going at the rate of 30 miles per

hour will be brought to rest in about 84 yards by continuous

brakes, if they press on the wheels with a force equal to ^ of

the weight of the train, the coefficient of friction being '16.

EXAMPLES ON CHAPTER II. 105

10. A train of 150 tons moving with a velocity of 50 miles

per hour has its steam shut off and the brakes applied and is

stopped in 303 yards. Supposing the resistance to its motion

to be uniform, find its value and find also the mechanical

work done by it measured in foot-pounds.

11. The driver of an express train whicli is running at

the rate of 70 miles per hour observes 660 yards ahead the

tail-lights of a goods train at rest. The brakes of the express

train, which exert a constant retardation, can pull it up in 500

yards when running with steam off at 50 miles per hour, and

the goods train can in one minute get up a velocity of 30 miles

per hour. If the express shuts off steam and puts on its

brakes and the goods train puts on full steam ahead, find the

respective velocities just before collision.

12. On a certain day half an inch of rain fell in 3 hours ;

assuming that the drops were indefinitely small and that the

terminal velocity was 10 feet per second, find the impulsive

pressure in tons per square mile consequent on their being

reduced to rest, assuming that the mass of a cubic foot of

water is 1000 ounces and that the rain was uniform and con-

tinuous.

13. Find the pressure in poundals per acre due to the

impact of a fall of rain of 3 inches in 24 hours, supposing the

rain to have a velocity due to falling freely 400 feet.

14. A cage of mass 2 tons is lifted in 2 minutes from

the bottom of a pit a quarter of a mile deep, by means of an

engine which exerts a constant tractive force, but the steam

is shut off during the ascent so that the cage just comes to

rest at the surface. Find the smallest possible horse-power of

the engine.


15. A train of mass 200 tons moving with a velocity of

21yy miles per hour comes to the top of an incline of 1 in 150,

and runs down with steam turned off for 1800 yards when it

comes to the foot of a similar incline;find the least horse-power

which will carry it to the top in one minute, supposing the

resistance due to friction etc., to be equal to 11 '2 Ibs. per ton.

16. A train of mass 200 tons is ascending an incline of 1

in 100 at the rate of 30 miles per hour, the resistance of the

rails being equal to the weight of 8 Ibs. per ton. The steam

being shut off and the brakes applied the train is stopped in a

quarter of a mile. Find the weight of the brake-van, the

coefficient of sliding friction of iron on iron being ^.

17. Assuming that the resistances of all kinds to the

motion of a train on the level are equal to 6 Ibs. per ton of the

total weight of the train when it is moving with less speed

than 10 miles per hour, with an addition of ^lb. per ton for

every mile of speed above 10, and that the total weight of the

train and the engine is 200 tons, and that the maximum speed

attainable is 40 miles per hour, find the time from rest of

reaching a velocity of 10 miles per hour, the engine always

exerting its full force.

18. A carriage is slipped from an express train going at

full speed at a distance I from a station;shew that, if the

carriage come to rest at the station itself, the rest of the train

Mwill then be at a distance -TT 1 beyond the station, M andM-mm being the masses of the whole train and of the carriage

slipped, and the pull exerted by the engine being constant.

19. From King's Cross to Grantham is 105 miles and there

are 27 intermediate stations. A parliamentary train stops at

all stations and the average resistance to its motion when the

brake is off is ir|^th of its weight whilst the resistance to an


express train which runs the distance without stopping is

5^-j-th of its weight, but, when the brake is on, the resistance

in eacli case is -^gth of its weight. Supposing the brake to be

always applied when the speed has been reduced to 30 miles

per hour and not before, find which train is most expensiveand by how much per cent.

20. A train of mass m moves against a uniform retardation

equal top. times its weight, and starts from rest moving with

uniform acceleration until the steam is shut off, and arrives at

the next station distant a from the starting point in time I.

Shew that the greatest horse-power exerted by the engine is

C a r- ,where C is a constant depending on the units

figt 2a

employed.

21. The resistance./? due to the air and friction to a train

moving at the rate of V miles per hour(V being > 10) is given

in pounds by the formula # =[6 + I (V- 10)] [T + 2M] where

E is the weight of the engine and T that of the train in tons.

If Fbe<10 then R= 6 (T+ 2E). Assuming this formula,

and that the force exerted by the engine is constant, that the

weight of the engine is 40 tons and that of the train 1 60 tons

and that 40 miles per hour is full speed for the train, find its

velocity at the end of 1 minute 24 seconds from the start.

22. An engine of horse-power H draws a train of M tons

up an incline of 1 in m, the resistance to motion apart from

gravity being equal to the weight of n Ibs. per ton. Shew that

the maximum speed is

550//mir/nftjir. T- feet per second.M (2240 + nni)

23. Two equal balls are let fall at the same instant, one

from the ridge of a house down the slates and the other from

the eaves. The balls roll so that when one touches the ground


the other is at the edge of the eaves, and the perpendicular to

the earth will pass through both balls;find the locus of the

centre of inertia of the balls during the motion.

24. Two equal masses, A and B, are connected by an

inelastic thread, 3 feet long, and are laid close together on

a smooth horizontal table 3J feet from its nearest edge ;B is

also connected by a stretched inelastic thread with an equal

mass C hanging over the edge. Find the velocity of the

masses when A begins to move and also when B arrives at the

edge of the table.

25. Two masses, each equal to W, are hung over a

smooth pulley, being connected by a string sixteen feet long,

and are arranged so that either on approaching to within two

2Wfeet from the pulley lifts a load

y=-. The motion is started

with one weight and its load at the pulley ;shew that the

motion repeats itself every fifteen seconds.

26. Two masses P, Q connected by a string passing over

& smooth pulley are both held at a distance c above a fixed

horizontal plane and motion ensues. If the plane destroy the

momentum of the heavier mass P as it hits the plane, shew

that the system comes to rest in time

27. A body of mass m Ibs. is attached to a weightless

inextensible string which passes over a smooth pulley, and is

attached at the other end to a body of mass m Ibs. lying on a

table, the line joining the position of m' to the pulley being

inclined at an angle a to the vertical. If the string be at first

slack, and become stretched when the body has fallen through

one foot, find the impulse of the tension and the velocity com-

municated to m'.


28. A mass M after falling freely through a feet begins to

raise a mass m greater than itself and connected with it bymeans of an inextensible string passing over a fixed pulley.

Shew that m will have returned to its original position at the

end of time

2M-, . /

9

2M/'2a-M V Q

'

29. An inelastic mass M hanging freely draws another

mass M' up a rough inclined plane by means of a string over

a pulley at the top of the plane. If M start from the top of

the plane and M' from the bottom, find the velocity of M'

whenM strikes the ground and, if p. be and the angle of incli-

y3nation of the plane be 30, find the subsequent motion.

30. At the bottom of a mine 275 feet deep there is an

iron cage containing coal weighing 14 cwt., the cage itself

weighing 4 cwt. 109 Ibs. and the wire rope that draws it G Ibs.

per yard. Find the work done when the load has been lifted

to the surface, and the horse-power of the engine required to

do that amount of work in forty seconds.

31. A man of 12 stone ascends a mountain 11000 feet

high in 7 hours, and the difficulties in his way are equivalent

to carrying a weight of 3 stone;one of Watt's horses could

pull him up the same height without impediments in 56

minutes; shew that the horse does as much work as 6 such

men in the same time.

32. A blacksmith wielding a 14-lb. sledge strikes an iron

bar 25 times per minute and brings the sledge to rest upon the

bar after each blow. If the velocity of the sledge on striking

the iron be 32 feet per second, compare the rate at which he

is working with a horse-power.


33. A train of 1 20 tons is to be taken from one station to

another, a mile off, up an incline of 1 in 80 in four minutes

without using brakes. Shew that neglecting passive resist-

ances the engine must exert a pull before the steam is turned

off equal to the weight of about 6203 pounds.

34. Find the loss of time and the extra expenditure of

work due to crossing a pass by a railway having an incline of

one in m up and one in n down, instead of going through a

level tunnel of length I through the pass, supposing that V is

the maximum speed allowed on the line and that the speed

drops from V at the bottom to v at the summit of the pass.

35 . Shew that the loss of time in going from A to C, two

points on a railway at the same level 8 miles apart, due to an

incline of 1 in 100 from A up to B and an incline of 1 in 300

from B down to (7, instead of going from A to C at a uniform

velocity of 45 miles per hour, is about 2 minutes 20 seconds.

It is supposed that with full steam on the velocity dropsfrom 45 miles an hour at A to 15 at the summit B, and that

in descending the incline from to C full steam is again kepton till the velocity has again reached 45 miles per hour, after

which the velocity is kept uniform by partly shutting off steam;and prove that this happens at a point Q distant from B 1 mile

892 yards.

36. If the mass of the train in the previous question be

200 tons, and the resistance due to friction etc. be 14 pounds

per ton, then the pull of the engine from A to Q is 2T5

tons,

and from Q to C, T7 of a ton

;find also the extra expenditure

of work due to the inclines.

37. In an Attwood's machine, if the string can only bear

a strain of one-fourth of the sum of the weights at the two

extremities, shew that the larger weight cannot be much less

than 6 times the smaller and that the least possible accelera-

qtion is -Z

.

EXAMPLES ON CHAPTER II. Ill

38. The locus of points from which the times down equally

rough inclined planes to a fixed point vary as the length of

the planes is a right circular cone.

39. Particles slide down rough chords of an ellipse from

the upper extremity of the major axis which is vertical. The

chord drawn to the extremity of the minor axis will be the

chord of quickest descent to the curve if the coefficient of

e*friction be

j-^ ,where e is the eccentricity of the ellipse.

-j *j 1 6

If the coefficient of friction be greater than this value, then

the major axis itself will be the chord of quickest descent.

40. Shew that points in a vertical plane from which a

particle will slide down a rough chord to a fixed point in a

given time lie on one or other of two fixed circles.

41. A pulley is suspended from a given point by a string;

over the pulley passes another string to whose extremities are

attached masses M and 3M. The masses are initially at rest;

after five seconds have elapsed the string supporting the pulley

is cut;find the distance of each mass from its initial position

at the end of five seconds more.

42. A mass m^ in descending draws up a mass m.2 bymeans of a string over a small pulley ;

if the latter after it has

been in motion 10 seconds pick up another mass m3 ,find the

total distance described by it in twenty seconds from the time

at which they start.

43. Two equal masses P, Q connected by a string over a

smooth pulley are moving with a common velocity, P descend-

ing and Q ascending. If P be suddenly stopped and instantlylet drop again find the time that elapses before the string is

again tight.


44. A string over a pulley supports a mass of 5 Ibs. on

one side and masses of 2 and 3 Ibs. on the other, the lower

mass 2 Ibs. being distant one foot from the other. The two-

pound mass is suddenly raised to the same level as the other

and kept from falling. Shew that the string will become taut

in half a second, and that the whole system will then movewith a uniform velocity of 3*2 ft. per sec.

45. A constant force acts on a particle at rest for a time

: it then ceases for time -, then acts for the same time,n nthen ceases, and so on alternately. Find the space described

in time t and the limiting value when n is infinite.

46. A heavy inelastic particle slides down an imperfectly

rough plane inclined at an angle 6 to the horizon; after de-

scribing a path, length a, it meets another equally rough planeinclined at an angle to the horizon up which it ascends

describing a path of length b before it comes to rest;shew that

0/1 sin (9-

e)b = a cos

2 26 . . } a r,sm (6 + e)

where e is the angle of friction when the particle is in motion.

47. A cylinder, of height h and diameter d, is standing on

the horizontal seat of a railway carriage. If the train begin

to move with acceleration f, shew that it will not remain

undisturbed unless f<p.g and also<-^, where//,

is the'i

coefficient of friction.

48. Two equal balls A and B are at a distance a apart.

A force of impulse / acts on A in the direction AB and a

constant force F on B in the same direction produced.Shew that A will not overtake B unless P > 2aFm, where ra

is the mass of either ball.


49. Corn flows uniformly through an opening in a floor

to a floor at a depth A below;shew that the centre of inertia

of the falling corn is at a distance 7. below the opening.o

50. Three particles are placed at the angular points of a

triangle ABC, their masses being proportional to the opposite

sides;shew that their centre of inertia is the centre of the

inscribed circle of the triangle, and that if they move alongthe perpendiculars to the opposite sides with velocities respec-

tively proportional to the cosines of the angles from which

they started, the centre of inertia will move in a straight line

towards the orthocentre.

51. A heavy particle is projected with velocity v up a

smooth inclined plane whose length and inclination are I and a

respectively. At the same instant the plane starts off with

uniform velocity ^. Shew that the particle will reach

the top in time -.

v

52. A heavy chain is drawn up by a given force P which

exceeds the weight, W, of the chain. Find its acceleration

and the tension at any assigned point.

L. D. 8

CHAPTER III.

THE LAWS OF MOTION (continued).

MISCELLANEOUS EXAMPLES.

[In a first reading of the subject the student may, with advantage,omit this Chapter.]

102. THE following chapter consists of miscellaneous

examples of a rather more difficult character than those

given in Chapter II.

In some we have systems of masses whose motions are

dependent on one another;in others the principles of the

Conservation of Energy and Conservation of Momentumas enunciated in Articles 95 and 101 are exemplified.

Ex. 1. A string passing over a smooth fixed pulley supports at its two

ends two smooth moveable pulleys of masses m1?m2 respectively. Over

each passes a string having masses mv m2 at its ends; shew that the

acceleration of each of the moveable pulleys is ~ ^ ;, andm^ 1001^2 +m2261

find the accelerations of the different masses, and the tensions of the

strings.

MISCELLANEOUS EXAMPLES. 115

Let A, B be the moveable pulleys of masses TOJ, my Let C, Ebe the masses m

1and D, F the masses ?/;.,.

Let T be the tension of the string round the

fixed pulley, and Tlt T2 the tensions of the

strings round the pulleys A and /', all three

tensions being expressed in poundals.

Let / be the acceleration of A downwards

or of B upwards.

and r-ro20-2ZT

2=ma /.

Hence solving for /, T we have

and

= 22^7712 + 222 m1+ 2i1

ms (1)

,2 (r,

-T,) m,

/= & + i

*D- m,

?g , (2) *C

Also the accelerations of C and D downwards are respectively

Tl

But since in the course of the motion the string AC lengthens as

much as AD shortens, therefore the motion of C relative to A must be

equal and opposite to the motion of D relative to A.

.'. acceleration of C - acceleration of A = -[ace. of D - ace. of A],

the accelerations being all measured downwards.

.-. ace. of C+ acc. of D= 2 . ace. of A.

Hence

and

Similarly ace. of E + ace. of F= 2 . ace. of B = -2f.

From (3), (4)

82

116 MOTION OF A WHEEL AND AXLE.

/. substituting in (2),

,_ _

f= 2% m^ + ?tt22

Substituting this value of / successively in equations (3) and (4) we

obtain T and T2 .

TAlso the acceleration of the mass C downwards=g ---

2/ma|

So the accelerations of the masses D, E, F are obtained.

Ex. 2. Tivo masses, M and Mj, balance on a wheel and axle ; if they

M (M M)be interchanged shew that the acceleration of Ma is ., 2 _ M^ irp 8

(/ie TTiass of the wheel and axle being neglected.

Let OA, OB the radii of the wheel and axle respectively be a, b so

that when there is equilibrium we have

aM=bM1 (1).

When the masses are interchanged as in the above figure, let T, Tl

be the tensions of the strings, and /, /i the accelerations of M, M^

respectively.

WEDGE WITH MASSES ON ITS FACES. 117

Then ^(M.g-T^M, .............................. (2),

f=(T-Mg)IM.................................. (3).

Now the resultant of T and 2\ must be balanced by the reaction at ;

for otherwise the forces on the machine will have a resultant, and this

would cause an infinite acceleration in the machine, since its mass is

negligible.T a H! ...

2^=6=^ ................................. <

4)'

Again since the machine turns round its centre, the distances moved

through by M, M1must be proportional to b and a, and so also their ac-

celerations.

Hence from (2), (3), (5) we have

M T-Mg Ml

M

(M l9- Tj =M? (T- Mg)=M^ H 2\

-Mg] , by (4).

...... L ^'-^ _J

Ex. 3. A smooth isosceles .wedge of mass M is placed on a smooth table

and carries a smooth pulley at its summit, and two masses rQj, m^ are

attached to the ends of a string which passes over the pulley ; shew

that the acceleration with which the string passes over the pulley is

mx- m, M + m, + m,- -

. v= ;

L. . , . g sin o, where a is the angle at the base ofm

: +m22 8

the wedge.Find also the other circumstances of the motion.


Let /!, /2 be the accelerations of % along and perpendicular to its

face of the wedge, /3 , /4 those of r 2 similarly, /6 that of the wedge hori-

zontally, and T the tension of the string. Then clearly

/mj, ............................. (1)

/m2 , ............................. (3)

/4=(E2 -m2fifCOs a)/?^, ........................... (4)

/5=(.R2 sin a- Bj sin a)/.if, ..................... (5).

[In this last equation the horizontal components of the tension of the

string vanish. Had the faces been inclined to the horizon at angles ctj,

02, there would have been an additional term T (cos c^- cos a2)

in the

numerator of (5) due to the difference of the horizontal components of

the tensions at the pulley.]

Also since the string is inextensible, the rate at which m.2 approachesA is the same as that at which m1 leaves A.

'

/3-/5 COS a= -/1-/6 COS a -

'. /i+/s=0 .................................. (6).

Again the accelerations of mv m2 respectively, perpendicular to the

faces, must be the same as the accelerations of the wedge in the same

directions ;for the masses are always in contact with the wedge.

.-. /2=/8 sina ................................. (7)

-/4=/B sino ................................. (8).

From (1) (3) and (6) we obtain

So from the remaining equations on solving

Bl _ E% _ g cos a

m^ (M+ 27 2 sin2 a)~m^ (M+ 2m1 sin2 a)

~ M + (m^ + wz2 )sin2 a

*

.-. acceleration with which the string slips over the pulley

7> _ -n

=/a ~/5 cos a=/3 - cos a sin a :

TO, TOo .M+Wi+TOo . , ,.= ^.

- ^--9^ 9 sin a > on reduction.TOJ+ TOJ HT+(TOj + TO2 )

sin a

22 .RAlso the acceleration of the wedge along the plane=/5

= sin a -2-,,-

1

-5 f/sin a cos a.

PERFORATION OF A PLATE BY A SHOT. 119

Also the pressure of the wedge on the plane

= Mg + (I?! +B2)cos a + IT sin a

sin2 a

,, sn a

4m1 ?>z sin2 a . (m, + wz,)2 sin2 a .- is < -- - is < (mi+m,,) sin2 a.Now

THj + H2 m1

Hence this pressure is < (M+ml +m2) g,

or the pressure is less than it would be if the string were fixed at some

point of its length so as to prevent motion and the wedge kept at rest.

Ex. 4. Assuming that in a cannon the force on the ball depends only on

the space occupied by the volume of the vapour of the gunpowder, shew

that the ratio of the final velocity of the ball when the gun is free to recoil

I Mto its velocity when the gun is fixed is . / ^ -, where M, m are the

masses of the cannon and ball respectively.

When the gun is free to recoil let the velocities of the ball and gun be

U, r initially.

Then since the total momentum generated in the gun in one direction

is equal to the momentum generated in the shot in the opposite direction,

we havemU=MV................................ (1).

Also since the change in the kinetic energy of a system is equal to the

work done on it, we have

^mtP +pfF^work done by the explosion ............... (2).

When the gun is fixed let Uj be the initial velocity of the shot. Then

^?;il712=work done by the explosion in the second case ............ (3).

Now since the pressure of the vapour of the powder depends only on

the space it occupies, and since in each case the powder expands from the

volume it occupied originally to the volume that the gas occupies in the

ordinary atmosphere, the work done in the two cases is the same.

and, substituting from (1) for V, we have

MM+m'


In this example the mass of the products of the explosion may be

either looked upon as negligible ;or it may be assumed that they are

expelled with the same velocity as the bullet and their mass included in

that of the bullet.

Ex. 5. If a be the penetration of a shot of m Ibs. striking a fixed iron

plate with velocity v, sheio that a plate of M Ibs. and thickness b, free

Mto move, will be perforated ifb< ^- a, and that after passing through

m+ ^/M2 -M(M + m) -

the plate the shot will retain the velocity ,, v, theM +mresistance being supposed uniform.

Let F be the resistance. Then since the shot is reduced to rest in

passing through a distance a of the plate, we have

%mv2=F . a .................................. (1).

Let x be the thickness of the thinnest plate, free to move, which will

just stop the bullet. When the distance x has been traversed the bullet

and plate are moving with a common velocity. Call this velocity U.

Then since the total momentum is unchanged by the impact, we have

mv = (m +M) U................................. (2).

Also since the loss of kinetic energy is equal to the work done we have

(3).

From (2) and (3) we have, by eliminating U,

MM+m'

Mso that the plate will be perforated if the thickness b be < .

If the shot pass through the plate, let its velocity on emergence be

V. The equations (2) and (3) now become

mv =mV+MU,and %mv

2 -\mV*- \MU*=Fb=%m- v3.

.. eliminating U we have

Mmv* - MmV2 - m? (v- Vy=Mm - v2

.

PILE-DRIVING. 121

Whence by solving

r= ,,M+m

Ex. 6. An inelastic pile of mans mlbs. is driven vertically a feet into

the ground by n blows of a hammer of M Ibs. falling vertically through h

feet ; shew that - - Ibs. superposed on the pile in addition to M would

drive it down very slowly supposing the resistance uniform.

If tJie pile be crushed x feet by each blow, where x is small, the mean

pressure exerted by the hammer is equal to the weight of ^ pounds,

and each blow lasts for r- of the time of falling of the hammer, neglecting

forces not due to the impulse.

Let F! be the velocity of the pile and hammer immediately after the

blow. Since the total momentum is unaltered by the blow,

M j2gh=(M+m)V1 ........................... (1).

Let F be the resistance of the ground. Then the total force on

the pile is F-(H+m)g, and this force brings the pile and hammer to

arest in a space

-.

/. } (M+m) V*=[F- (M+m) g] .-

.

n

--.2 a M+m

.: the resultant upward pressure = weight of -- pounds so thatM+mathis mass superimposed would just overcome the resistance and force the

pile slowly downwards.

Again during the impact let P denote the mean pressure between the

hammer and the pile. Then since the change in the kinetic energy of the

hammer is equivalent to the work done, we have

M . 2</ft- (M+m) V?- Px.

Substituting for F, from (1), we have P = ^m -g,M+mx

. Mm hor mean pressure = weight of - pounds.M +mx


Also time of each blow

_ change in the momentum of the hammerP

_ M J2gh- MV1 _ M(M+m) Jfyh-M* J~~Mm a

x

x /2h x= / = - x time of falling of the hammer.n V g h

Ex. 7. Two men, each of mass M, stand on two inelastic platforms,

each of mass m, hanging over a small smooth pulley. One of the men

leaping from the ground would be able to raise his centre of inertia through

a distance h; shew that if he leap with the same energy from his platform

his centre of inertia would rise a distance ^^-r h, and that he would& (JM -j- mj

fall on the platform in its original position and the whole system be again

at rest.

Since the man can raise his centre of inertia through a height h, the

work he does on his system by leaping must be Mgh.

Let U, V be the initial velocities of the platform and the man.

Then since the momentum of the man is equal and opposite to that

of the system consisting of the platforms and the other man,

.-. MV=(M+2m)U .......... ................. (1).

Also since he leaps with the same energy in the two cases we have

^M7 2+ ^(M+2m)U 2=Mgh....................... (2).

Solving these equations we have

-^ -M+m a

Hence height through which his centre of inertia rises

F2 M+2m2.g 2 (M+m)

h.

VAlso time before he returns to his original position= 2 (3).

After he has left the platform it starts with initial velocity U downwards

Mand moves with an acceleration -r~ q in the opposite direction.M+ 2m

MOTION OF AN ELASTIC RING ON A CONE. 123

.-. if T be the time before the platform is in its original position

again, we have

Hence the man and his platform are in their original positions again

at the same instant, and since their momenta are then equal and opposite

the whole system comes to rest.

Ex. 8. A heavy elastic ring in the form of a circle is placed with its

plane horizontal over a smooth cone whose axis is vertical; if the ring be

held in contact with the cone at its natural length and allowed to fall, how

far will it descend before first coming to rest ?

Since the change in the kinetic energy of a system is equal to the

work done on it, and since the kinetic energy of the ring both at the

beginning and end of the motion is zero and hence the total change zero,

the total amount of the work done on the ring during the motion must be

zero also.

Let 2a be the vertical angle of the cone, 2ira the unstretched length of

the ring, X its coefficient of elasticity, and let 2irx be the length of the

ring when it first comes to rest.

The work done by the weight of the ring during the motion is

W(x -a) cot a, where W is its weight.

Also the work done by the tension is (Art. 91, Ex. 2),

-i (2irx

-2ira) \ ^*

,or -

(x-

a)2

.

Also the work done by the normal pressure of the cone is zero.

(Art. 89.)

Hence we have W (x-a) cot a-- (x

-a)

2= 0.

JPacota , ,. . , Wa.: x-a=-

; ,and distance the string descends = - cotjo.

T\ X IT

Ex. 9. A goods train consists of a number of similar waggons and an

engine whose mass is an integral number(/JL)

times the mass of a waggon.

They are coupled by chains of equal length, inelastic, and weightless. The

train is initially at rest on a straight line of railway with all the vehicles

in contact and the couplings slack. The engine then begins to move, the


steam exerting a constant tractive force, and each waggon starting with a

jerk as its coupling tightens. Shew that, during the starting, the velocity

of the moving part of the train will be greatest just before the nth impact,where n = 2/*

a-3/* + l, provided there are at least this number of impacts.

All rotation of the loheels and friction may be neglected.

Let Vn be the velocity of the train just before the ?ith truck starts, Units velocity just after the nth truck starts.

Then since the total momentum of the train is unaltered by the jerkwhich sets the nth truck in motion, we have

(/t + n-l)Fs= (M + n)l7n ........................ (1).

Also if F be the force in poundals exerted by the engine and m the

mass of a truck, the acceleration when n trucks are in motion is

Hence since the train moves for a distance a with this acceleration,

where a is the length of a coupling. Hence from (1) (2)

So

But V1= velocity of the engine just before the first truck moves

fan

.-. adding and substituting for V1 we have

2Fa 2Fa~m M=

So ra_--- FaF -m

~IJ"Hence the greatest value of Vn is when

~" -1= 0, or when

MOTION OF HEAVY STRINGS. 1 I*.")

103. We shall now solve some questions relating to

the motion of heavy strings. The best method for a

student to obtain a clear idea of the motion is to consider

the case of a light string on which are fixed a number of

masses, equal to one another, and then to consider the

limiting case when the distances between the particles

become indefinitely small. In this manner we pass to

the case of a uniform heavy string. If the string be not

of uniform density, but have its density given accordingto some known law, then the masses should be so chosen

that ultimately they may form a string of the required

density. For example, if the heavy string be given to be

of density which varies as its distance from one end, the

masses should be chosen so as to be in the ratio of 1, 2,

3, ...; if it vary as the square of the distance from one

end, the masses should be chosen in the ratios of I2,22

,

3s,

. . .; similarly any other case may be considered.

When a string is in motion there being no jerks (as in

the case of a string stretched tight on a table and drawn

over the edge by a portion which hangs vertically) the

principle of the Conservation of Energy may be applied.

Ex. 10. A heavy uniform string is coiled up on the edge of a horizontal

table and a small portion hangs over the edge ; if motion be allowed to

ensue, shew that the end descends with uniform acceleration g, and find

the tension at the edge at any time.

Consider the case of a light string laden with equal masses m at equalintervals a.

Let Vn, Un+1 be respectively the velocities of the system just before

and after the (n+ l)th particle comes over the edge.

Since the total momentum of the system is unaltered by the jerk

necessary to set the (n + l)th particle in motion,

n+1=nVn (1).

126 MOTION OF HEAVY STRINGS.

Also between successive jerks the system moves freelywith acceleration g,

.: Vn*=Un'+ 2ga (2).

Hence from (1) and (2)

So n2 [^2_

(_

1)2 Un_^_ 2 a .

(n _ i)2 >

Therefore by addition

3 n+ 1

But Ul= initial velocity of the first mass=0.

2_ ga

Now let na= x, so that .

3 x + a

If we make a very small we have the case of a uniform heavy chain so

that if Ux denote the velocity when a distance x of this heavy chain has

passed over the table we have

V*=12x=

2

-j-x, (by neglecting a).

Hence the end of the chain is descending with a constant accele-

qration | .

o

Let M be the mass of the string which is hanging vertically at anyinstant and T the tension at the edge of the table.

Then Mg - T= moving force on this portion=M .

o

.: T= =two thirds of the weight of the portion which hangsB

vertically.

Ex. 11. One end of a heavy uniform string hangs over a small pulleyand the other end is coiled up on a table; if the part hanging over the

pulley be initially of length b and the distance of the pulley from the table

be c (< b) and motion ensue, find the velocity at any time and the tension

of the string at the table.

Consider the case of a light string with masses m attached to it at

equal intervals a; initially let p of these be hanging on one side and q on

the other.

MOTION OF HEAVY STRINGS. 127

Let Vw Un+1 be respectively the velocities of the system just before

and after the (n + l)th mass passes over the pulley ; also let us supposethe distance a so chosen that as one mass passes over the pulley another

is lifted from the table.

Then since the total momentum of the system is unaltered by the

jerk necessary to set this particle in motion,

.-. (n+p + 1) U^^n+p) Vn...................... (1).Now when n particles are on one side of the pulley, and p on the other

tn _jj

the acceleration of the system= g, and for a distance a the systemj

is moving with this acceleration.

F 2=^ +2a ........................... (2).

From (1), (2) we have

(n +p + 1)^ U^ -(n+p? tfn

o

So n+ 2 U*

(q +p + I)2 [7a+1

2 -(q +PY Uf =2ga {q* -p*}.

Also Uq , being the initial velocity of the system, is zero.

Hence, by addition,

(n+p + 1)2 Un+1*=2ga[q*+(q + 1)2+ ... + n2 - (n- q + l)p*\,

_

Now put na= x, pa=c, qa=b.Then this equation is

If we now make n indefinitely great and a indefinitely small we obtain

the case of the question in which we have a uniform heavy chain.

If Ux be the velocity when a distance x - b has passed over the pulley

we obtain

(x + c)2 Ux*= 2gp^ - c2 (x

-6)]

=^ (x-

ft) [*2+ xb+ 62 - 3c2].

Let T be the tension of the string at the table. Then the impulse of

this tension in a small time r is TT. Also in this time a portion of the

string equal to m UXT is put into motion, where m is the mass of the

string per unit of its length ; also its velocity is changed from to Ux ;

hence the momentum generated is mf7z2r. But the impulse of a force is

equivalent to the momentum generated, so that T=mUx*.

128 TRANSMISSION OF POWER.

104. Transmission of power by belts and shafts.

When a moving belt passes over a pulley, the work done

by the belt on the pulley can be easily found. For let

TI}T

2be the tensions of the two portions of the belt at

the points where it meets and leaves the pulley respec-

tively ;.also let v be the velocity of any point of the belt

in feet per second. Then the work done by the tension

T3is Tj) and the work done against the tension T^ is Tj}.

Hence the total work done on the pulley per second is

As an example, let us find the horse-power transmitted to a shaft bya band which travels at the rate of 44 feet per second round a pulley

which is firmly keyed to the shaft, when the tensions of the two portions

of the band arc 600 and 200 Ibs. weight respectively.

Here r2=6000; ^ = 200*;; v=44.

.*. work done per second=400 g . 44 = 400 x 44 foot-pounds.

400 x 44Hence the horse-power transmitted = =32.

550

105. When a couple of given moment is applied to a

shaft which is turning with a given angular velocity the

horse-power transmitted is known.

Let &> be the angular velocity of the shaft, G the

moment of the couple applied to it, and r the radius of

the section of the shaft.

The couple G may be represented by two forces =-Zfl*

acting at the ends of a diameter. In the unit of time the

point of application of each force moves through a distance

C1 C1

rw. Hence the work done by each force is _ . ra> or^-&>,

and therefore the work done by the couple is G . &>.

EXAMPLES ON CHAPTER III. 129

Hence the work done in one second is - - foot pounds9

and the horse-power transmitted is Go* + 550 g.

For example, if a shaft be acted on by a couple of 10000 foot-pounds

and is revolving 50 times per minute the work done on it per minute is

10000 x 100?r foot-pounds.

10000 x lOOa-Hence the horse-power transmitted IB

ooUUU

EXAMPLES. CHAPTER III.

1. In a system of pulleys (three moveable) in which all

the strings are attached to the roof, the highest string after

passing over a fixed pulley has a mass m attached to it, and

the lowest pulley a mass M. If M descend shew that its

acceleration is (M 8m)g/(M+ 64m).

How is this expression altered if each pulley have a mass /* ?

2. In the system of pulleys where each string is attached

to a bar which supports the weight, if there be two moveable

pulleys, and the power be quadrupled, the weight will ascend

with acceleration ~ .

2u

3. In a system of pulleys with n strings at the lower

block the upward acceleration due to a power P is

(nP-W)g/(n*P + W).

If, when W has an upward velocity v, the weight P reach

the ground there will presently be on the string a strain of

impulse nPWv/(irP+ W).

L. D. 9

130 EXAMPLES ON CHAPTER III.

4. Two weights, P and W, equilibrate on a wheel and

axle, the mass of which may be neglected. A weight W is

attached to P and after the lapse of one second another

weight W is attached to the ascending weight W ;shew that

after the lapse of another second the velocity of the ascending

weight 2W is -=-^ = ^rff, where a, b are the radii of thea2 + ab + 262 *

wheel and axle respectively.

5. A fine string passes over a smooth fixed pulley, and

carries at its ends two small smooth pulleys, each of mass one

pound, and these in turn carry strings with masses 1 lb., 2 Ibs.

and 1 lb., 3 Ibs. tied to their ends respectively. Shew that

the acceleration of each of the moveable pulleys is -^ .

2io

6. A string one end of which is fixed has slung on it a

mass m^ and then passes over a fixed pulley having a mass m<,

attached to it at the other end ;find the accelerations of the

system and the tension of the string.

7. A string carries at one end a mass M, and at the

other a pulley of mass M', and is placed over a fixed pulley;

over the pulley M' passes a string carrying at its ends masses

m and m'; shew that the acceleration of M4mm' + (M' M} (m! + m)~4mm' + (M' + M) (m' + m)

^'

Compare the spaces passed over by M and m, in a given time,

and shew that the ratio of the upper and lower tensions is

4mm' + M' (m + m') : 2mm'.

8. A string passes over a fixed pulley ,and has a mass

m at one end and a pulley, C, of mass p, at the other. Astring is fastened to a fixed point A, vertically below B, passes

over (7, and has at its other end a mass m';shew that the

acceleration of the pulley is (2m'- m +p) g/(m + 4m' +p).

EXAMPLES ON CHAFfER III. 131

9. A smootli ring of mass M is threaded upon a string

whose ends are then placed over two smooth fixed pulleys

with masses m, m tied at the other ends respectively, and the

various portions of the strings hang vertically. Find the

motion of the ring and the two weights, and shew that the

ring will be at rest if M= 4mm'/(m + m').

Explain how the question is changed if M be fixed on the

string.

10. One end of a string is fixed;

it then passes under a

moveable pulley to which a weight W is attached, and then

over a fixed pulley and a smaller weight P is attached to its

other end, all three sections of string being vertical; shew

that, neglecting the mass of the pulley, the acceleration with

which W descends is(W - 2P) g/(W -

4P).

11. Two pulleys whose weights are W and W are

connected by a string hanging over a smooth pulley; over

the former is hung a string with weights P, Q at its ex-

tremities and over the latter one with weights R, S. Shew that

when the system is free to move the acceleration of the pulleys

W-W + 2H-2H' . , pis = =-, jrjf njj, g, where H is the harmonic mean ot P,

Q and H' that of A1

,S.

12. Over a pulley is placed a string at one end of which

is a single weight; at the other end is another pulley over

which is slung a string carrying masses m, m' at its ends;

find the magnitude of the single weight so that m' mayremain at rest, if initially so, and shew that the acceleration

of the pulley is - -g, if its weight be neglected.

_ 1 1 (

13. The masses of three particles are m, 2m and 3m; mand 2m are joined by a string passing round a pulley and

placed on a smooth table to the edge of which the strings are

92


perpendicular; 3m is joined by a string to the pulley, supposed

weightless, and hangs over the edge ;shew that the acceleration

of the mass 3m is

14. A weight Wlis placed on a rough table and has tied

to it a light string which passes over the edge of the table and

supports a pulley of weight W2 ;round this pulley hangs

another light string which has attached to it two unequal

weights W3 ,TF4 ;

find the acceleration of the different parts of

the system and also the relation between the different weights

and the coefficient of friction that ]\\ may remain at rest.

15. A sailor of weight W sits in a loop of rope attached

to the lower block of weight W of that system of pulleys in

which one rope passes round all the pulleys. The lower block

is suspended by n vertical chords. If P' be the force the

sailor must exert on the free end of the rope in order to

maintain equilibrium, shew that if he exert a force P he will

begin to move with an acceleration (n+ l)(P-

P') g/(W + W).

16. In any machine without friction and inertia a weight

P supports a weight both hanging by vertical strings ;if these

weights be replaced by P and W, and if in the subsequent

motion P and W move vertically, the centre of inertia of Pand W will descend with acceleration

[WP'-PWJ

17. Two equal weights P are attached to the ends of a

string, six parts of which are vertical, and which passes over

three fixed pulleys and under two moveable pulleys each of

weight P ;another string has its ends fastened to the centre

of the two moveable pulleys and supports a pulley and weight

of joint weight 2P. When the whole is in equilibrium a

downward blow is given to one of the two weights. Compare

the velocities communicated to the various pulleys and weights.


18. 2n small smooth rings are fixed at equal intervals in

a horizontal circle and an endless string passed through them

in order. If the loops of the string between each consecutive

pair of rings support pulleys of masses P, Q, R ... respectively,

the portions of string not in contact with the pulleys being

vertical, shew that the pulley P will descend with acceleration

1 1 \ /I 1 1

19. A bullet of mass one ounce is fired from a rifle

weighing 6 Ibs. freely suspended and the initial velocity of the

bullet is found to be 1000 feet per second. Find what the

initial velocity of the bullet would be if the rifle were rigidly

held.

20. Two bullets of the same size but compounded of

metals whose densities are in the ratio 3 : 4 are shot

vertically upwards in a vacuum at the same instant from

two similar guns charged with the same quantities of powder;one returns to the point of projection half a minute later than

the other;find the velocities of projection.

21. A shell of mass M is moving with velocity V. Aninternal explosion generates an amount E of energy and

thereby breaks the shell into masses whose ratio is m^ to w2 .

The fragments continue to move in the original line of motion

of the shell;shew that their velocities are

V.Tr tynJEand V- t / -

-If-jr.m9M.

22. A bullet fired horizontally from a musket being

supposed to pass perpendicularly through a target suspended

freely in the air; explain on the dynamical theory of work

why the greater the velocity of the bullet the less the

displacement of the target.


23. Shew that the resistance of wood is 204 Ibs. weight to

a nail of mass one ounce, supposing that a hammer of mass

1 Ib. striking it with a velocity of 34 feet per second drives

the nail one inch into a fixed block of wood.

If the block be free to move and be of mass 68 Ibs., shew

that the hammer will drive the nail only ^iths of an inch.

24. Two equal saucers lie on a smooth horizontal table

and a perfectly rough insect jumps from the centre of one to

the centre of the other and then back to the centre of the

first. Shew that the final velocities of the saucers are in the

ratio M + m : M, where M is the mass of each saucer and mthe mass of the insect.

25. Two buckets of given length are suspended by a fine

inelastic string placed over a smooth pulley; at the centre of

the base of one of the buckets a frog of given mass is sitting ;

at an instant of instantaneous rest of the buckets the frog

leaps vertically upwards so as just to arrive at the level of the

rim of the bucket;find the ratio of the absolute length of the

frog's vertical ascent in space to the length of the bucket

and shew that the time which elapses before the frog again

arrives at the base of the bucket is independent of the frog's

weight.

26. Two masses m, m' are connected by an elastic string

and are placed on a smooth horizontal table. Initially they

are at rest and the string unstretched;a blow, whose impulse

is P, is then given to the particle m along the direction of the

string ;shew that when the string is next unstretched the

2/>velocity of m' is

,and find the velocity of m at the samem + m

instant.

27. Two particles A, B of equal mass are connected by a

rigid rod of negligible mass, and a third equal particle C is


tied to a point P on the rod at distances a, b from the two

ends. C is projected with velocity u from P perpendicular to

the rod. Shew that just after the string becomes tight, C1 a3 + b2

will move with velocity ^ ; u, and find the initial2 a2 + ab + 62

velocities of A and B.

28. Three equal particles are placed at the angular pointsof a triangle ABC and A, B are connected with C by equal

strings which are just taut C is projected perpendicularlytowards AB. Find the loss of kinetic energy after AB are

jerked into motion.

29. A wedge of mass M supports two masses m, n one on

each face and they are connected by a string passing over the

summit and the whole is placed on a smooth table;shew that

if during the motion the string break then the acceleration

of M will be instantaneously reversed in direction providedmsina ... cos/8 ,

: lie between and unity.n sm p cos a

SO. A shaft is transmitting a couple equal to 2000 foot-

pounds and is making 150 revolutions per minute. WhatH.-P. does it transmit ?

31. A shaft is driven by an engine through the inter-

vention of a belt which travels at the rate of 30 feet per

second, the tensions of the free portions of the belt being

equal to the weights of 50 and 100 Ibs. respectively. Whatis the horse-power transmitted ?

32. A heavy wheel 5 feet in radius revolves on its axle

which is horizontal and two inches in diameter making 10

turns per minute, its mass being supposed to be entirely

collected at its rim. If left to itself the coefficient of friction

being \, how many turns will it make before stopping? How

136.

EXAMPLES ON CHAPTER III.

many turns will it make if it initially were making 20 revolu-

tions per minute ?

33. A uniform rectangular block ABCD stands on its

base AD on a rough floor. It is pulled at C by a horizontal

force just great enough to begin to turn it round the corner

D. If the same force continue to pull horizontally at C till

the block has turned through an angle 9 and then cease to

act, shew that the block will just have acquired sufficient

momentum to cause it to turn over round D provided that

sin 9 = tan - where a is the angle BDC.

34. A heavy elastic ring in the form of a circle is placed

with its plane horizontal over a smooth sphere whose axis is

vertical;

if the string be held in contact with the sphere at

its natural length and allowed to fall, how far will it descend

before first coming to rest? What is the condition that it

should just slip over the sphere 1

35. Two fixed points in the same horizontal line are

connected by an elastic string whose natural length is the

distance between the points and whose coefficient of elasticity

IV

-jj ',

a particle of weight w is fastened to the middle point of

\/3

the string. If the particle be held so that the string is

unstretched and be let go, shew that the angle 20 between the

two strings when the particle has fallen to its greatest depth

is given by the real root of the equation 4 sin3 9 + 3 sin = 1.

36. Two strings each of length a are fastened to a

particle of mass m. The one end of one string is fastened to

a fixed point A, and the other passes over a small smooth

pulley B in the same horizontal plane with A and carries a

particle of mass m. The distance AB is equal to a. If m, be


originally at rest and very near,the string being tight, shew

that just before m reaches B the velocity of m' is

m - 2m''

2What happens if m be < ^ m' ?

37. Two equal masses, each equal to P, balance by means

of a string connecting them which passes over two pulleys, A,B in the same horizontal line. If a third mass M be placed at

C, the middle point of AB, it will descend until

BAG WTort

)

"~

2P'

Shew also that if a be the value of the angle BAG at anytime the downward velocity of the mass M will be

f_ Jftana-2P(seca-l)l*} 2cicf ^

}

where AB=2a.

38. A railway train consisting of a number of carriages,

is so coupled that there is a distance of, say, three feet

between each carriage and the next. Explain why the

resistance which such a train experiences from a cross-wind

should be greater than from a head-wind when the train is

going at full speed.

If the transverse section of the carriages be a square of

7 feet side, the velocity of the wind 45 miles per hour, the

velocity of the train 60 miles per hour, and there be 10

spaces each 3 feet wide between the carriages, shew that the

resistance due to the wind is equal to the weight of more than

a ton, the mass of a cubic foot of air being taken to be l ozs.

39. Two particles, each of mass M, are tied one at each

end of an inelastic string of length 2, and a particle of mass


in is tied at its middle point. The system is laid on a smooth

horizontal plane with the string at its full length in a straight

line. A blow, of impulse F, perpendicular to the string is

given to the middle particle. Shew by means of the equations

of Energy and Momentum that, when the two portions of the

string each make an angle with their initial positions, the

Pangular velocity of either is- .

40. A wedge, whose section is a right-angled triangle,

rests on a rough table with the hypotenuse inclined to the

table at an angle i. A heavy smooth chain of length a is

fastened to the wedge near the right angle, and passing over

a smooth pulley hangs over the edge of the table, so that the

wedge is just on the point of motion in the direction of the

pulley when a length b of the chain hangs over the table. Aheavy particle is then placed on the inclined face of the wedgeand slides down it. Shew that the particle will separate from

the wedge when the end of the chain has descended through a

distance(

- + ajcot i, where

/*,is the coefficient of friction

between the wedge and the table, the motion being supposedto take place in a vertical plane.

41. An umbrella, whose surface is smooth and spherical,

is held in rain which is falling vertically with velocity v.

The umbrella itself is being drawn vertically downwards with

velocity F. Prove that, if V be less than v, the average

pressure per unit area of the rain falling on the umbrella at a

point whose angular distance from the highest point is 0, is

(v-F)2 cos2

,

p i-'--,where p is the average pressure per unit area

of the rain falling on a fixed horizontal plane.

42. A large number of equal particles are fastened at

unequal intervals to a fine string, and then collected into a


heap at the edge of a smooth horizontal table with the

extreme one just hanging over the edge. The intervals are

such that the times between successive particles being carried

over the edge are equal ;shew that if cn be the interval

between the nth and (n + l)th, and vn be the velocity just after/ M

the (n + I )th particle is carried over, .then = = n.Cl Vi

43. n equal masses are placed in contact in a line on a

smooth horizontal table, each being connected with the next

by an inelastic string of length a, and another equal mass is

attached to the foremost of these masses by a string which

passes over a pulley at the edge of the table. Shew that, of

the kinetic energy generated until the last mass is set in/n

motion the fraction - =-r- is lost, supposing none of the2 (n + 1)

masses to leave the table until the last is set in motion.

44. A fine inelastic thread is loaded with n equal

particles at equal distances c from one another;

the thread

is stretched and placed on a smooth horizontal table perpen-dicular to its edge over which one particle just hangs; shew

that the velocity of the system when the rth particle is just

/ r(r-l)leaving the plane is . / gc .

\/ 7l

Hence shew, that if a heavy string of length a be similarly

placed on a smooth table, the velocity with which it leaves the

table is Jag.

45. n particles are attached to a string at equal distances ;

it is placed so that r of the particles are on an inclined plane

and n r hang vertically ;the system is set in motion by

drawing one of the particles just over;shew that the velocity

of the system when the last particle just leaves the plane is

*J(rI)gc, where c is the length of the string between con-

secutive particles.


46. A number of equal particles are fastened at equaldistances a on a string and placed in contact in a vertical

line; shew that if the lowest be then allowed to drop, the

velocity with which the nth begins to move is

/ (n-l)(2n-l)>

\ 3n

47. An indefinite quantity of a uniform chain is coiled in

a heap on the floor of a room, and escapes into the room

below through a hole in the floor;shew that the velocity of

escape can never exceed Jya, where a is the height of the

hole above the floor of the room below.

48. A piece of uniform chain hangs vertically from its

upper end, the lower end just touching a smooth inclined

plane initially ;if it be let go shew that at any instant during

the motion frds of the pressure on the plane is due to the

impacts.

49. A fine uniform string of length 2 is in equilibrium

passing over a small smooth pulley and is just displacedshew that the velocity of the string when just leaving the

pulley is

50. A chain of length a is coiled up on a ledge at the

top of a rough inclined plane, and one end is allowed to slide

down. If the inclination of the plane is double the angle of

friction A, the chain will be moving freely at the end of time

6a cot A.

51. A fine uniform chain is collected into a heap on a

horizontal table;to one end is attached a fine string which,

passing over a small smooth pulley vertically above the chain,

carries a weight equal to that of a length a of the chain;


shew that the length of the chain raised before the weightcomes to rest is a J3, and find the subsequent motion.

52. A large number of equal particles are attached at

equal intervals, a, to a fine string which passes through a short

fine semi-circular tube, and initially 2r particles are on one

side, the highest being at the tube, and r particles on the

other, the lowest being in contact with a horizontal table

where the other particles are heaped up ;shew that the

velocity just before the nih additional particle is set in motion

/nag (n-1

}

V 3 1" (w + 3r-l)aj'l)

a

and deduce the corresponding result for a uniform chain

hanging over a pulley.

53. A uniform heavy chain is vertical, both ends beingfixed. The upper end A is then released and the chain falls

past the lower end B without striking it. At the instant

that A is passing ,the latter is released. Shew that the

chain will become straight again in of the time in which Afell to B.

54. A uniform heavy chain is coiled up in a man's hand

and one end tied to a fixed point. If the hand be suddenly

removed, shew that the strain on the fixed point at any time

during the motion is equal to three times the weight of the

portion of chain hanging vertically at that time.

55. A length I of uniform chain is lying in a heap on a

horizontal plane, and a man takes hold of one end and

continues to raise that end in a vertical line with uniform

velocity v. Shew that the force exerted by the man is

greater than the weight of the vertical portion at any time in

the ratio k + x : x, where x is the vertical portion and h twice

the height to which the velocity v is due.


56. A uniform flexible chain of indefinite length, the

mass of the unit of length of which is m, lies coiled on the

ground, whilst another portion of the same chain forms a coil

on a platform at a height h above the ground, the intermediate

portion passing round the barrel of a windlass placed above

the second coil. An engine which can do H units of work

per unit of time is employed to wind up the chain from the

ground and to let it fall into the upper coil. Shew that the

velocity of the chain can never exceed the value of v given by

the equation mghv + ^mv3 H.

57. A chain, whose density varies as the distance from

one end, is coiled up close to the edge of a smooth table and

the end allowed to hang over. Shew that the motion is

uniformly accelerated, and that the tension at the edge of the

table varies as the 4th power of the time that has elapsed

since the commencement of the motion.

CHAPTER IV.

UNITS AND DIMENSIONS.

106. WHEN we wish to state the magnitude of anyconcrete quantity we express it in terms of some unit of the

same kind as itself, and we have to state, (1), what is the

unit we are employing, and, (2), what is the ratio of the

quantity we are considering to that unit. This latter

ratio is called the measure of the quantity in terms of

the unit. Thus if we wish to express the height of a

man we may say that it is six feet. Here a foot is the

unit and six is the measure. We might as well have said

that he is 2 yards or 72 inches high.

The measure will vary according to the unit we

employ. The measure of a certain quantity multipliedinto the unit employed is always the same (e. g.

2 yards= 6 feet = 72 inches).

Hence if k, k' be the measures of a physical quantitywhen the units used are denoted by [K], [K'], we have

and hence [K] : [K'} : : :,

so that, by the definition of variation, we have [K] oc -=- or/c

144 UNITS AND DIMENSIONS.

the unit in which any quantity is measured varies inverselyas the measure and conversely.

107. A straight line possesses length only and no

breadth or thickness, and hence is said to be of one

dimension in length.

An area possesses both length and breadth, but no

thickness, and is said to be of two dimensions in length.The unit of area usually employed is that whose lengthand breadth are respectively equal to the unit of length.

Hence if we have two different units of length in the

ratio X : 1, the two corresponding units of area are in the

ratio X2: 1, so that if [A] denote the unit of area and

[Z] the unit of length, then

A volume possesses length, breadth, and thickness, and

is said to be of three dimensions in length. The unit is

that volume whose length, breadth, and thickness are each

equal to the unit of length. As in the case of areas it

follows that if [ F] denote the unit of volume then

Since the units of area and volume depend on that of

length they are said to be derived units, whilst the unit

of length is called a fundamental unit.

Another fundamental unit is the unit of time usuallydenoted by [T]. A period of time is of one dimension in

time.

The third fundamental unit is the unit of mass denoted

by [Jfj. Any mass is said to be of one dimension in mass.

These are the three fundamental units;

all other units

depend on these three, and are therefore derived units.

UNIT OF VELOCITY. 145

108. Theorem. To shew that the unit of velocity

varies directly as the unit of length, and inversely as the

unit of time.

In one system let the units of length, time, and velocity

be denoted by [L], [T], and [V], and in a second system by

[L'],[T], [V]; also let

[L']= m[L], and [T']

= n[T].

Then a body is said to be moving

with the original unit of velocity

when it describes a length [L] in time [T] ;

.'. with velocity m [V~\

when it describes a length m [L] in time [T] ;

.'. with velocity [F]

when it describes a length m [L] in time n [T] ;

771

/. with velocity [F]

when it describes a length [L'] in time [T'].

But it is moving with velocity [V] when it describes

a length [L'] in time [T].

m.

/. by the definition of variation, [7] oc LJ .

L. D. 10

146 UNIT OF ACCELERATION.

109. Theorem. To shew that the unit of acceleration

varies directly as the unit of length, and inversely as the

square of the unit of time.

Take the units of length and time as before, and let

[F], [F] denote the corresponding units of acceleration.

Then a body is said to be moving

with the original unit of acceleration

when a vel. of [L] per [T] is added on per [T],

.: with acceleration m[F]when a vel. of ra [L] per [T] is added on per [T],

.-. with acceleration [I7

]n L

when a vel. of ra [L] per n [T] is added on per [T],

.: with acceleration ,[F]n

when a vel. of ra [L] per n [T] is added on per n [T],

/. with acceleration ~*[F]

when a vel. of [L'] per [T] is added on per \T].

But now the body is moving with the new unit of accele-

ration

.-. [Ff

] : [F] :: ra : ri>

..1*3.in[L] [TY

'

[T]2 '

EXAMPLES. 147

Ex. 1. Find the measure in the centimetre-minute system of the

acceleration due to gravity, taking its measure in the foot-second system to

be 32-2, and assuming a metre to be 39'37 inches.

In a falling body a velocity of 32 '2 ft. per sec. is added on per sec.,

60x32-2 ft. ... min per sec.,

. 602 x32-2 ft. ... min per min.,

GO2 x 32-2 x 12

3937"cms. per mm per min.,

. , 3600x12x32-2 QMao ,

.-. required measure =57r5=- = 3533249.

"

This may be more concisely put as follows :

Let x be the new measure ;

..*[**] = 32-2 [JF].

,- 32 .2 xX

12= 32-2 x -s^- x 602,as before.

*

Ex. 2. Convert a horse-power into C.G.S. units, assuming as a rough

approximation that 1000 grammes= 2 Ibs. and 1 centimetre= '4 inch.

We have 1 Ib. = TBT . 10s grammes, and 1 ft. =30 cms.

Also weight of 1 gramme = 981 dynes roughly.

Now the agent is dragging

the weight of 33000 Ibs. through 1 foot in 1 min.,

. ............... 550 Ibs.......... 1 foot ... 1 sec.,

. ................ 30x550 Ibs.......... 1 cm. ... 1 sec.,

30 . 550 . 5 . 10s. ................ - -= grms.......... 1 cm. ... 1 sec.,

.. the agent drags a force equal to ~ x 981 dynes through

1 cm. per sec. and is therefore doing about 7'4 x 109 c. o. s. units of

work per second.

Ex. 3. If the unit of length be one mile, and the unit of time one

minute, find the units of velocity and acceleration.

Ans. 88 feet per second ; f foot-second units.

102

148 DIMENSIONS.

Ex. 4. If the unit of velocity be a velocity of 30 miles per hour, and

the unit of time be one minute, find the units of length and acceleration.

Ans. 880 yards ; }\ foot-second units.

Ex. 5. If the unit of mass be 1 cwt., the unit of force the weight of

one ton, and the unit of length one mile, shew that the unit of time is

^/SS seconds.

110. Dimensions. Def. When we say that the dimen-

sions of a physical quantity are a, & 7 in length, time, and

mass respectively, we mean that the unit in terms of which

the quantity is measured varies as

Thus the results of Arts. 108, 109 are expressed by

saying that the dimensions of the unit of velocity are 1 in

length and 1 in time;while those of the unit of accele-

ration are 1 in length and 2 in time.

The cases in Arts. 108, 109 have been fully written

out, but the results may be obtained more simply as in

the following article.

111. (1) Velocity. Let v denote the numerical

measure of the velocity of a point which undergoes a

displacement whose numerical measure is s, in a time

whose numerical measure is t, so that

s = vt.

If [L] [T] [F] denote the units of length, time, and

velocity respectively, we have111s oc

r-y= ;t x ==

;vx

[L]' [T]' [F]'

JL 1 *

'[]* mm'

UNITS AND DIMENSIONS. 149

(2) Acceleration. Let v denote the velocity acquired

by a particle moving with acceleration f for time t so that

v=ft.

If [F] denote the unit of acceleration we have

1 11<X

(3) Density. Let d be the density of a body whose

mass is m and volume u so that

m = dii.

If [D], [ U] denote the units of density and volume we

have'

1 1x; * x[Tfj'

If the body be very thin so that it may be considered

as a. surface only, we see similarly that the unit of surface

density

So if the body be such that its breadth and thickness maybe neglected, (so that it is a material line only) we have

unit of linear density oc [M] [L]^1

.

150 UNITS AND DIMENSIONS.

(4) Force. If p be the force that would produceaccelerationf in mass m we have

p = mf.

.'. if [P] denote the unit of force we have

(5) Momentum. If k be the momentum of a mass

m moving with velocity v we have

k = mv.

.'. if [JT] denote the unit of momentum,

(6) Impulse. If i be the impulse of a force p acting

for time t we havei = pt.

.'. if [/] denote the unit of impulse,

so that an impulse is of the same dimensions as a

momentum.

(7) Kinetic Energy. If e be the kinetic energy of

a mass m moving with velocity v we have

e = | mv*.

.'. if [E] denote the unit of kinetic energy,

(8) Work. If w be the work done when a force pmoves its point of application through a distance s then

w =ps.

.'. if [W] denote the unit of work,

Hence work and kinetic energy are of the same dimensions.

EXAMPLES. 151

(9) Power or Rate of work. If h be the power at

which work w is done in time t, then

i w *-ih = =wt .

t

.'. if [IT] denote the unit of power,

EXAMPLES.

1. If the acceleration of a falling body be taken as the unit of accele-

ration, and the velocity generated in a falling body in one minute as the

unit of velocity, find the unit of length.

In ft.-sec. units the velocity generated in one minute= 60 x 32.

Let [L], [T] denote a foot and second respectively, and [L'], [T'] the new

units of length and time.

Then since the dimensions of an acceleration are [L] [T]~2

, we have

l.[L'][r']-2 = 32[L][T]-

2......................... (1).

Also, since the dimensions of a velocity are [L] [T]"1

, we have

l.[L'][T']-1 = 60.32[L][I

T

]-1..................... (2),

.'. dividing the square of equation (2) by (1) we have

.-. the new unit of length= 32 x 602 feet.

2. The kinetic energy of a body expressed in the foot-pound-second

system is 1000 ; find its value in the metre-gramme-minute system having

given 1 /oot = 30'5 cms. ; 1 Ib. = 450 grammes approximately.

Let x be the measure in the new system, so that

['] = 1000 [E],

or x [AT] [I/]2[I"]-

2= 1000 [M] [L]2[T]-

2.

But [Jf]=450[aT], [L]=-305[Z/], [T-\

.: *=1000x450x(-305)2 x60a

= 150,700,500.

152 EXAMPLES.

3. If the kinetic energy of a train of mass 100 tons and moving at tJie

rate of 45 miles per hour be represented by 11, while the impulse of the

force required to bring it to rest is denoted by 5, and 40 horse power by 15,

find the units of time, length, and mass, and shew that the acceleration due

to gravity is denoted by 2016, assuming its measure in foot-second units to

be 32.

In foot-second units the kinetic energy of the train= . 100 . 2240. 662.

The impulse of the force required to bring to rest=100 . 2240 . 66.

Also 40 H.-P. = 40 x 550 x 32 units of work per sec.

Hence since the dimensions of energy are [M] [L]2[T]~

2,

11 [M'] [L']2[r']-

2=i . 103 . 224 . 662[Jf] [I/pfT]-"......... (1).

Also since the dimensions of an impulse are [M] [L] [TJ"1,

.-. 5 [!T][LT[T7-i = 108.224. 66 [M][L][T]~l............ (2).

Similarly considering the dimensions of a power,

15 [AT] [L']2[rj-

3=102. 55 . 128[M][L]

Z[T]~

3............ (3).

Dividing (1) by (3),

'

[T'] = 21 . 45 seconds= 15| minutes.

Also dividing (1) by (2), V [L'] [r']-i = 33 . [L][T]-i.

.-. [Z/] = [L] . 15 x 21 . 45= 105 x 135 feet=2|^ miles.

Also substituting in (2),

Hence the acceleration due to gravity

= 32 212 453=2016 new units -

4. If the unit of time be one minute, the unit of mass 500 pounds, and

the unit of density that of water, find the unit of length, assuming that the

mass of a cubic foot of water is a thousand ounces ; shew also that the unit

offorce is about ^ of a poundal.

The new unit of density is i Ibs. per cubic ft.

But

EXAMPLES. 153

[L'][n-2_ 5

[L] [I']-2"

5. Given that the unit of power is a million ergs per minute, that the

unit offorce is a thousand dynes, and the unit of time one-tenth of a second,

what are the units of mass and length?

Let [M], [L], [T] denote a gramme, a centimetre and a sec., and [M'],

[L'], [2"] the new units.

10^Since a million ergs per minute = -^ ergs per second

oU

108= -s-original units of power,

o

105

.-. l.[af][LT[2*r3= ~[M][L]*[T]-3............... (1).

Similarly since the new unit force= 103 . old unit force,

.-. l.[M'][L'][T']-2=103

[J^[L][:ri-2.................. (2).

Also [F] =Am .................................... (3).

102

Dividing (1) by (2) we have [L'] [T']-1 =

=J-[L] [T]~\

102.-. by (3), [L']=

-y-[L] .A= | centimetre.

Also substituting in (2),

[M'] .-| [L] . 10a [r]-

2= 103 [M] [L] [T]-2

.

.: [M'] = 6 [M] = 6 grammes.

/. the required units are 6 grammes and f cm. respectively.

6. A certain physical quantity is represented algebraically by a single

term. If the unit of length be doubled the measure is \ of its former value,

whilst if the units of length and time be both doubled the measure is also

doubled. If in the former case the unit of mass be defined as the mass of

unit volume of some substance, the measure is -^nd of its former value.

What kind of quantity is it ?

Let o, /3, 7 be the dimensions of the quantity in mass, length, and

time ; and let y be its measure when expressed in terms of the original

unit [F] and let [FJ, [Y2], [F3] be the units in the three following cases so

that

154 EXAMPLES.

Let [M '] be the unit of mass in the last case. Then

m : [MT [Lfm* = = [rj : IMT&Lf m?:: [FJ : [M]

a[2Lf [2T]* :: [T3] :[M']

a[2Lf [2'f ... (2),

Now in the last case the unit of mass is denned as the mass of unit

volume; and since the unit length has been altered in the ratio 1 : 2, the

unit volume has been altered in the ratio 1:8;

.-. [M']= 8[M] ................................. (3).

Hence substituting from (2) in (1) and using (3), we have

l=i.2/J =2.20.2y=

1&.8a.2

/3

,

.-.solving /3=2; 7 = -3; a= l.

so that the physical quantity is a power or rate of doing work.

7. Taking as a rough approximation 1 foot = 30-5 cms.; 1 lb. = 453

grammes; and the acceleration of a falling body = 32 ft. -sec. units, shewthat

(i)1 Poundal = 13816 Dynes,

(ii) 1 Foot-Poundal = 421403 Ergs,

(iii) 1 Erg = 7-416 x 10~8Foot-Pounds,

(iv) 1 Horse-Power =7-416 x 109 Ergs per sec.,

(v) 1 Horse-Power = 7'6 x 106 gm.-cms. per sec. ,

(vi) 1 Foot-Pound= 13816 gm.-cms.

8. If the unit of distance be one mile and the unit of time 4 seconds

find the units of velocity and acceleration.

Ans. 1320 ft.-sec. units; 330 ft. -sec. units.

9. If the unit of acceleration be that of a freely falling body, and the

unit of time be 5 seconds, shew that the unit of velocity is a velocity of

160 ft. per sec.

10. What must be the unit of space, if the acceleration due to gravity

be represented by 14, and the unit of time be five seconds ?

.4n8. 57| feet.

11. The acceleration produced by gravity being 32 in ft.-sec. units,

find its measure when the units are TTTT^TJ of an hour and a centimetre;

given 1 centimetre '0328 ft.

Ans. 126|f.

EXAMPLES. 155

12. If the area of a ten acre field be represented by 100, and the

acceleration of a heavy falling particle by 58$, find the unit of time.

Aw. 11 seconds.

13. If the unit of velocity be a velocity of 3 miles per hour, and the

unit of time one minute, find the unit of length.

Ans. 88 yards.

14. If the acceleration of a falling body be the unit of acceleration

and the velocity acquired by it in 5 seconds be the unit of velocity, shew

that the units of length and time are 800 feet and 5 seconds respectively.

15. If a hundredweight be the unit of mass, a minute the unit of

time, and the unit of force the weight of a pound, find the unit of length.

Ans. 342^ yards.

16. If the units of velocity, length, and force be each doubled the

units of time and mass will be unaltered, and that of energy increased

in the ratio 1 : 4.

17. If the unit of work be that done in lifting one hundredweight

through three yards, and the unit of momentum that of a mass of one

pound which has fallen vertically 4 feet under gravity, and the unit of

acceleration three times that produced by gravity, find the units of time,

length, and mass.

Ans. 21 sees. ;14112 yards ; ^ Ib.

18. If 5^ yards be the unit of length, a velocity of one yard per

second the unit of velocity, and 6 poundals the unit of force, what is the

unit of mass ?

Aiis. lllbs.

19. If the unit of time be one hour and the units of mass and force

be the mass of one hundredweight, and the weight of a pound respectively,

find the units of work and momentum in absolute units.

Ans. The unit of work=f X1204 absolute units of work; the unit

of momentum= 8 x 1202 absolute units.

20. If the unit of force be the weight of 8 pounds, whilst a cubic foot

of a substance of unit density is of mass 81 Ibs., and the unit of time is

one second, shew that the unit of length is 1 foot 4 inches.

156 VERIFICATION OF FORMULAE

21. If the units of force, work, and time be taken as the fundamental

units, what are the dimensions of the unit of length?

Ans. 1 in work and - 1 in force.

22. If the unit of force be the weight of one pound, what must be

the unit of mass so that the equation P=mf may still be true ?

Ans. g pounds.

Verification of formulae by means of countingthe dimensions.

112. Many formulae and results may be tested bymeans of the dimensions of the quantities involved.

Suppose we have an equation between any number of

physical quantities. Then the sum of the dimensions in

each term of one side of the equation in mass, length,

and time respectively must be equal to the corresponding

sums on the other side of the equation. For supposethat the dimensions in length of one side of the equationdiffered from the corresponding dimensions on the other

side of the equation ;then on altering the unit of length

the two members of the equation would be altered in

different ratios and would be no longer equal; this however

would be clearly absurd; for two quantities which are

equal must have the same measures whatever (the same)unit is used. For example, if two sums of money are

the same, their measures must be the same whether we

express the amounts in pounds, shillings, or pence.

113. Much information may be often easily obtained

by considering the dimensions of the quantities involved.

Thus the time of oscillation of a simple pendulum (which

consists of a mass ra tied by means of a light string of

BY MEANS OF THEIR DIMENSIONS. 157

length I to a fixed point) may be easily shewn to vary as

-. For, assuming the time of oscillation to be inde-

pendent of the arc of oscillation, the only quantities that

can appear in the answer are m, I, and g. Let us assume

the time of oscillation to vary as nfvff*.

The dimensions of this quantity expressed in the usual

way are f [T,'] )>

\M]a

[L\*\

or

Now the answer is necessarily of one dimension in time,

and none in mass or length. Hence we have

= 0; -27 = 1.

^illation x A / -

V gand the time of oscillatior

'V g

114. As another example, consider the case of a body

which moves in a straight line with an acceleration ^LCL

toward a fixed point in the straight line, where d is the

distance of the moving body from that fixed point.

Now ^ is an acceleration and must therefore be of onea

dimension in length and minus two in time.

.'./j,must be of 3 dimensions in length and 2 in time.

We can easily shew that the time the body takes to

move to the centre of force, starting from rest at a distance

a from it, must vary as * /

158 ATTRACTION UNITS.

For the only quantities that can appear in the answer

are /*, and a. Let us assume then that this time

oc fjf . ay.

The dimensions of this latter quantity are then

{[L]3

[T]-'2

}

x.[L]

y or [L]Sx+y

[T]-*x

.

Hence, as before, 2x = 1, and 3x + y = ;

.-. x = -% and y = \,

:. the required time oc/j,~^ a?

or asv>*

Ex. 1. If a body be projected with Telocity V from a point in a

horizontal plane, shew that, neglecting the resistance of the air, the range

on the horizontal plane mustV*

oc .

g

Ex. 2. From a consideration of dimensions only shew that if a bodybe acted on by a constant force in the direction of its motion, the space

passed over in time t from rest must vary as the square of the time.

Would the same considerations enable us to find the space when there is

an initial velocity ?

Ex. 3. A particle moves from rest under a central force equal to

if T be the time from rest at distance a to the centre of force,distance

'

shew, by the consideration of dimensions only, that

115. Attraction units. The law of gravitation

states that every particle in nature attracts every other

particle with a force which varies directly as the product

of the masses of the particles, and inversely as the square

of the distance between them. Hence the force of attrac-

ATTRACTION UNITS. 159

tion between two particles of masses mltm

a , placed at a

771 JYL

distance r apart, is X,*

a

2 where X is some constant

quantity.

This expression for the mutual attraction of two

particles can be simplified by properly choosing either the

unit of mass or the unit of force.

The unit of mass can be so chosen that the constant X

may be unity and the attraction between the two masses

expressed in dynes. This may be done by considering

the attraction of the earth on a particle at its surface.

Let the mass of the earth be E grammes, and let the

new unit of mass be x grammes, so that the mass of the

f 1earth is new units of mass, and one gramme is - new

3) \K

units of mass.

Then the attraction between the earth and one

E I

cc cc

gramme is ^ dynes where R is the radius of the earth

in centimetres.

But the attraction of the earth on one gramme= weight of one gramme = 981 dynes.

E 1

Hence = 981.

Now E = 6-14 x 1027

;R = 6'37 x 108

.

r^~.'. x = A/ - ^ = 3928 approximately.

160 ATTRACTION UNITS.

Hence the unit of mass, when the attraction between

two masses is V^ dynes, is 3928 grammes approximately.

The corresponding unit of mass belonging to the

/ /<*

Foot-Pound-Second system is A/ --p-2 pounds, where E isV g . H

the number of Ibs. in the earth's mass, R is the number

of feet in the radius of the earth, and g is equal to 32

approximately.

116. We can, if we wish, simplify the expression for the attraction

by keeping a gramme as the unit of mass and choosing a new unit

of force.

In this case the new unit of force will be the attraction between two

masses, each one gramme, placed at a distance of one centimetre from

one another. Hence the attraction between the earth and a gramme on

E .1its surface will be ^2

- units of force. But this attraction is 981 dynes.

981 Rz

Hence the unit of force = =j dynes.M

But, as in the previous article, J2= 6'14 x 1027; J?=6-37 x 108.

Hence we easily see that this unit of force would be 6 '48 x 10~8dynes.

117. The constant X can be so determined that the masses are

expressed in grammes and the attraction between them in dynes. In

this case the attraction between the earth and a mass of one grammeon its surface is

E 1

EHence X.--

The attraction between two masses, of m1and m2 grammes respectively,

placed at a distance of r centimetres is therefore 6-48 x 10~ 8 x ~dynes.

ASTRONOMICAL UNIT OF MASS. 161

118. Astronomical unit of mass. Instead of usinga pound or a gramme as the unit of mass we can simplify

the expression for the attraction between two bodies by

choosing as our unit mass that mass which would attract

an equal mass with the corresponding unit of force;this

unit of force is the force which in the new unit of mass

produces unit acceleration.

Thus, using centimetre-second units, let x grammes be

the new unit of mass. Then, as in Art. 115,

E 1

pa- = weight of one gramme

= weight of - units of mass = x 981,x x

Eso that # = ----

^2= 1-543 x 10

7

approximately.

This unit is called the Astronomical Unit of Mass of the

centimetre-second system, and is equal to T543 x 107

grammes.The attraction between two masses m

l ,m

2 ,at a

77? 77i

distance r apart, is' 2

. This attraction is, therefore,

of dimensions [.M]8

[Z]~*. But a force is of dimensions

mmm*.Hence [M]* [L]~*

= [M] [L] [T]~z

.

Hence the astronomical unit of mass is of 3 dimensions

in length and 2 in time.

The astronomical unit of density is of dimensions

[M] [L]~3

or of dimensions [T7

]"2

. It is therefore inde-

pendent of the unit of length.

L. D. 11

162 UNITS AND PHYSICAL CONSTANTS.

TABLE OP DIMENSIONS AND VALUES OP FUNDAMENTAL

QUANTITIES.

[Many of the following values are taken from Prof. Everett's

Units and Physical Constants.]

Dimensions in

Physical Quantity Mass Space Time

Volume density 1 3

Surface density 1 2

Line density 1 1

Velocity 1 1

Acceleration 1

Force 1 1 -2

Momentum 1 1 - 1

Impulse 1 1 1

Kinetic energy 1 2 - 2

Power or Rate of work 1 2 3

Values of "g."

Place Ft.-sec. units Cm.-sec. units

The equator 32-091 978-10

Cape of Good Hope 32-140 979-62

Latitude 45 32-17 980-61

Paris 32-180 980-94

London 32-19 981-17

North Pole 32-252 983-11

The value of g at any place is approximately given by

g = g<> [i_ -00257 cos 2A - 1 '96 x h x 10~ 9

],

where g is the value of gravity in latitude 45 at the sea level,

X the latitude of the place and h its height above the sea level

in cms.

UNITS AND PHYSICAL CONSTANTS. . 163

1 cm. = -39370 inches = -032809 feet.

1 foot = 30-4797 cms.

1 gramme = 15-432 grains = -0022046 Ib.

1 Ib. = 453 '59 grammes.

1 dyne = weight offf^T gramme approx.

1 poundal = 13825 dynes.

1 foot-poundal = 421390 ergs.

1 foot-pound = 1-356 x 107

ergs.

1 foot-pound = 13825 gramme-centimetres.1 erg = 7-37 x 10~8

foot-pounds.

1 Joule = 107

ergs = about foot-pound.

1 Horse-power = 7-604 x 106

grms-cms. per sec.

= 7*46 x 109

ergs per sec.

= 746 Watts.

1 Force de Cheval =7-5 x 10" grms-cms. per sec.

1 Watt = 1 Joule per sec.

= 107

ergs per sec.

Radius of the earth = 6 '37 x 10" cms.

Mean density of earth = 5-67 grammes per cm.

Mass of earth = 6-14 x 10" grammes.

Radius of the moon = 1-74 x 108 cms.

Mean density of moon = 3-6 grammes per cm.

Mass of the moon = 6*98 x 1025

grammes.

Mean distance between centres of moon and earth

= 3-84 x 10 10cms.

Mass of the sun = 324000 x that of earth.

Mean distance of sun and earth = 1-487 x 10 13 cms.

= 9-239 x 107 miles.

Velocity of a point on the equator of the earth about its axis

= 46510 cms. per sec.

Mean velocity of the earth in its orbit

= 2960600 cms. per sec.

112

164 EXAMPLES ON CHAPTER IV.

EXAMPLES. CHAPTER IV.

1. If the unit of acceleration be that of a body falling

freely, the unit of velocity the velocity acquired by the bodyin a seconds from rest, and the unit of momentum that of one

pound after falling for b seconds; find the units of length,

time, and mass.

2. Find the units of mass, length, and time supposingthat when a force equal to the weight of a gramme acts on

the mass of 16 grammes the acceleration produced is the unit

of acceleration, that the work done in the first four seconds is

the unit of work, and that the force is doing work at unit

rate when the body is moving at the rate of 90 cms. per

second.

3. In a certain system of absolute units the acceleration

produced by gravity in a body falling freely is denoted by 3,

the kinetic energy of a 600 pound shot moving with velocity

1600 feet per second is denoted by 100, and its momentum by

10; find the units of length, mass, and time.

4. If the unit of velocity be a velocity of one mile per

minute, the unit of acceleration the acceleration with which

this velocity would be acquired in 5 minutes, and the unit of

force equal to the weight of half a ton, find the units of length,

time, and mass.

5. The velocity of a train running 60 miles per hour is

denoted by 8, the resistance the train experiences and which

is equal to the weight of 1600 Ibs. is denoted by 10, and the

number of units of work done by the engine per mile by 10.

Find the units of mass, length, and time.

EXAMPLES ON CHAPTER IV. 165

6. In two different systems of units an acceleration is

represented by the same number whilst a velocity is repre-

sented by numbers in the ratio 1:3; compare the units of

tii! ic .and space.

If further the momentum of a body be represented bynumbers in the ratio 5 : 2 compare the units of mass.

7. The units of mass, energy, and distance being taken

as fundamental, find the dimensions of the unit of time in

terms of these units.

8. What is the number of Ibs. in the unit of mass, if the

attraction between two particles of masses m^ m respectively

YTL 77i

placed at a distance of r feet be ^2

poundals, assuming the

earth to be a sphere of 4000 miles radius and of mean density5 '67 times that of water?

9. Assuming the attraction between two masses m^ mz

771- Wito be A.

.,

2

,shew that, when ft.-lb.-sec. units are used, the

value of X is approximately 1-02 x 10~ 9.

[Use the values given in the preceding question.]

10. If the unit of force be defined as the attraction

between two unit masses at unit distance, find the unit of

mass in pounds. [The units of space and time are taken to

be one foot and one second respectively, and gravity to be due

to the attraction of a sphere of 4000 miles radius and havinga specific gravity of 6, that of water being 1.]

Find also the attraction of two Ibs. mass which are a yard

apart in terms of the weight of one pound.

11. Assuming the attraction between two masses m1 , m^

to be X r-2, when absolute units are used, find the dimensionsrof X.

166 EXAMPLES ON CHAPTER IV.

12. Find in dynes the attraction of gravitation of two

homogeneous spheres, each of mass 10 kilogrammes, the

distance between their centres being a metre; given that a

quadrant of the earth supposed spherical is 109 centimetres

the mean density of the earth 5-67 grammes per centimetre,

and the acceleration due to gravity 981 c. o. s. units.

13. Taking the value of gravity as 981 in the centimetre-

sec, system and the earth's radius as 6 -37 x 108centimetres,

find the earth's mass in astronomical units.

14. Shew that the attraction between the moon and the

earth is about 2-169 x 1028dynes having given the following

data;moon's distance = 60 x earth's distance

;moon's radius =

1740 kilometres; moon's density = 3 -6 grammes per centimetre,

and acceleration due to gravity on the earth = 982-8 C.G.S. units.

CHAPTER V.

PROJECTILES.

119. IN the previous chapters we have considered onlymotion in straight lines. In the present chapter we shall

consider the motion of a particle projected into the air

with any direction and velocity. We shall suppose the

motion to be within such a moderate distance of the

earth's surface that the acceleration due to gravity maybe considered to remain sensibly constant. We shall also

neglect the resistance of the air and consider the motion

to be in vacuo; for, firstly, the law of resistance of the air

to the motion of a particle is not accurately known, and,

secondly, even if this law were known, the discussion

would require a much larger range of knowledge of puremathematics than the reader of the present book is sup-

posed to possess.

120. Before proceeding further it may be advisable to state some of

the properties of the curve called the parabola which are most -useful

for our purpose. We shall give no proofs. They may be found in

any book on Geometrical Conies such as those by Dr Taylor, Dr Besant,or the Rev. .W. H. Drew.

168 PROPERTIES OF THE PARABOLA.

(1) A parabola is the path traced out by a point which movesso that its distance from a fixed point S is the same as its perpendicular

distance from a fixed straight line XM, so that if P be any point on

the parabola then SP=PM.

(2) The point S is called the focus, XM the directrix, and the line

SX, perpendicular to XM, the axis of the curve. The point A where the

axis meets the curve is called the vertex.

(3) The double ordinate LSL' through the focus perpendicularto the axis is called the latus rectum and is equal to 2SX or AS.

(4) If PN be the perpendicular from any point on the axis, then

PN2=4AS.AN.

(5) The tangent PT at P makes equal angles with the axis and the

focal distance SP, and therefore SP= ST; also AT=AN.

(6) If PG be the normal, then NO is equal to the semi-latus

rectum.

(7) If SY be drawn perpendicular to the tangent at any point P,then Y lies on the tangent at the vertex A and SY2=AS . SP.

PROPERTIES OF THE PARABOLA. 169

(8) A line through any point P of the curve parallel to the axis

is called the diameter through P and any chord QVQ' drawn parallel

to the tangent at P is an ordinate to the diameter PV.

(9) If QQ' be any ordinate to the diameter PV, then the tangents at

Q, Q' meet PV in a point T such that TP= PV, the line QQ' is bisected

in V, and QV*~4SP . PV.

Conversely if we can shew that the square of the distance QV,

drawn from a variable point Q in a fixed direction to meet a fixed line

PV in V, is proportional to the distance intercepted between V and

a fixed point P, it follows by a reductio ad absurdum proof that Q lies

on a parabola whose axis is parallel to PV and whose tangent at Pis parallel to Q V.

121. Def. When a particle is projected into the air

the angle that the direction in which it is projected makes

with the horizontal plane through the point of projection

is called the angle of projection ;the path which the

particle describes is called its trajectory; and the dis-

tance between the point of projection and the point where

the path meets any plane drawn through the point of

projection is its range on the plane.

170 THE PATH OF A PROJECTILE

122. Theorem. A particle is projected into the air

with a given velocity and angle of projection ; to shew that

its path is a parabola.

Let P be the point of projection, u the velocity and

a the angle of projection TPac.

The motion will take place in the vertical plane pass-

ing through PT ;for there is nothing to take the particle

out of this plane, the only force which acts on the particle,

viz. its weight, being in this plane.

Consider the position of the particle at the end of anytime t ;

measure off PT along the direction of projection

equal to ut. Draw TQ vertically downwards and equal to

\gt\

Then at the end of time t the particle would, if no

force acted on it, be at T\ but its weight gives it an

IS A PARABOLA. 171

acceleration g in a direction vertically downwards and

therefore would, in time t, draw it through a distance

Hence, by the principle of the Physical Independence

of Forces, (Art. 79), the particle is at Q at the end of

time t.

Draw QV parallel to PT to meet the vertical line

through P in V.

Then QV = PT = ut, and PV = TQ = \gt\o,,

.-. for all values of t, QV*= iff = .PV.Is

Hence if a parabola be drawn passing through P,

having PV as a diameter, touching PT at P, and such

2u2

that, if 8 be its focus, then 4P =,

it follows by Art.

\j

120 (9) that Q is always on this parabola.

123. The velocity of a projectile at any point of its pathis equal, in magnitude, to the velocity that the projectile

would acquire in falling freely from the directrix to that

point.

For let KX be the directrix of the parabola and draw

PK perpendicular to it.

Then, as in the last article,

Hence u is the velocity that would be acquired bya particle in falling through a vertical distance PK.

(Art. 51.)

But any point on the curve may be considered as a

starting point for the subsequent portion of the path ;for

the particle after passing through any point of its path

172 HORIZONTAL AND VERTICAL COMPONENTS.

moves in the same way as it would if projected from that

point with the velocity which it then has in the direction

in which it is then moving. Hence the proposition is

true for every point on the path of the projectile.

Cor. From this proposition it follows that the heightof the directrix above any point of the path depends only

on the magnitude of the velocity at that point and not on

the direction. Hence if a number of particles be pro-

jected with the same velocity from the same point in the*

same vertical plane but at different inclinations to the

horizon, their paths have a common directrix, and the foci

of their paths lie on a circle whose centre is the point from

which the particles are projected.

124. Instead of considering the motion in two direc-

tions which are respectively parallel to the direction of

motion and vertical, we shall generally find it more con-

venient to separately consider the motion in the horizontal

and vertical directions.

The initial horizontal velocity is u cos a and, since

there is no force acting on the particle in the horizontal

direction, there is no change in the horizontal velocity.

The initial vertical velocity is u sin a and the vertical

acceleration is g. Hence the vertical motion of the

particle is the same as that of a particle starting with

initial velocity u sin a and moving with acceleration g,

and the horizontal motion of the particle is the same as

that of a particle moving uniformly with constant velocity

u cos a.

The motion is the same as that of a particle projected with velocity

u sin a inside a smooth vertical tube of small bore, whilst the tube

at the same time moves parallel to itself with constant velocity u cos a.

GREATEST HEIGHT AND RANGE. 173

1 25. To find the greatest height attained by a projectile.

Let A be the highest point of the path, and let the

time of describing the arc PA be T.

Then T is the time in which the initial vertical

velocity is destroyed by gravity.

Hence, by Art. 41, = u sin a gT.

rp_u sin a

~7~'giving the time to the highest point of the path. Also, bythe same article, = ws

sin2 a 2g . AN.

4*0it sin. OL

.'. AN =9 , giving the greatest height attained.*Q

126. To find the range on the horizontal plane throughthe point ofprojection and to determine when it is greatest.

When the particle arrives at P the distance it has

described in a vertical direction is zero. Hence, if t be

the time of flight, we have

= u sin at $gt?.

2w sin a.". t ^ .

9

But during this time t the horizontal velocity has re-

mained constant and equal to u cos a.

/. PP = horizontal distance described in time t

2u* sin a cos a= u cos at =9

Hence the range is equal to twice the product of the

initial vertical and horizontal velocities divided by g.

Again with a given velocity of projection the range is

a maximum when sin 2<z is greatest and then a = 45.

174 TWO DIRECTIONS OF PROJECTION.

127. From the preceding article it may be shewn

that, in general, there are for a given range two directions

of projection which are equally inclined to the direction of

greatest range.

For let 6 be the required angle of projection when the

range and velocity of projection are K and u respectively,

, u2sin 20

Then = K.9

/. 811120 =^.w2

Now, if gK be not greater than w2

, there are two

values of 20, both less than 180, for each of which the

sine has the same magnitude, and these two values are

supplementary. Hence, if 6l , 2

be the correspondingvalues of 0, we have

Hence the two directions of projection PTV PTZare

T,

P

equally inclined to the direction PT which gives the

maximum range.

LATUS-RECTUM. 175

128. To find the lotus-rectum of the path described.

Consider the motion at the highest point A of the

path. The vertical velocity here is just zero and the

particle is moving horizontally with the constant hori-

zontal velocity u cos a.

But the velocity at A is that due to a fall through the

distance AX.

. 2w2cos

2 a.. latus-rectum = 2oJL = 4<AX =-= twice the square of the horizontal velocity divided

by-It will be noted that the latus-rectum, and therefore

the size of the parabola, is independent of the initial

vertical velocity and depends only on the horizontal

velocity.

Cor. The height of the focus above the horizontal

line through Pu2

sin2

a. u* cos2 a w2

= AN-AS = = - = -^-00820.2# 2# 2g

Hence, if a be less than 45, the focus of the path is

situated below the horizontal line drawn through the

point of projection.

129. To find the velocity and direction of motion after

a given time.

Let U be the velocity and ^ the angle which the

direction of motion at the end of time t makes with the

horizontal.

176 VELOCITY AT ANY POINT OF PATH.

Then 7 cos ^ = horizontal velocity at this time

= u cos a, the constant horizontal velocity.

And U sin-^r

the vertical velocity at this time

= u sin a gt.

Hence, by squaring and adding,

jij... wsina atand, by division, tan *lr = .

ucosa.

Cor. From this article we can at once deduce the important propo-

sition in Art. 123. For the height of the particle at time t above its

original starting point = u sin a . t - %gtz

.

it2

/. its depth below the directrix at time t= -(u sin a . t - %gt

2),

^9

and the square of the velocity acquired in falling through this distance

= 2g x ~- - u sin a . t + %gp~\ = u? - 2ug t sin a +gH2= U2.

130. Given the velocity and direction of projection, to

construct the focus and directrix of the path.

Let P be the point of projection and PT the direction

of projection, so that PT is the tangent to the path at P.

u*Draw PK vertical and equal to ^- ,

so that K is a*9

point on the directrix and a horizontal line through K is

the directrix.

Draw PS on the other side of PT from PK so that

the angle TP8 is equal to the angle TPK. Then, byArt. 120 (5), it follows that the focus must lie on the line

PS.

Make PS equal to PK;then S is clearly the focus.

EQUATION TO PATH. 177

131. To find the equation to the path of the projectile.

Take as axes of coordinates the horizontal and vertical

lines through the point of projection, P. Let Q be the

position of the particle at the end of time t, and let (x, y)

be its coordinates.

Then y is the vertical distance described in time t bya particle starting with velocity u sin a. and moving with

acceleration g.

.'. y u sin a . t

So x = u cos a . t.

Hence, eliminating t, we have

2 u' cos* a

This equation can be expressed in the form

2w2cos

2 ax sin a cos a = y _-

9 1. A 2

Hence the path is a parabola whose vertex is the point

/w2

. w2sin

2a\ , . 2w2

cos2a

sin a cos a,-

. whose latus-rectum is

\9 fy J g

and whose axis is vertical and concavity turned down-

wards.

EXAMPLES.

1. A bullet is projected with an initial velocity of 1280 feet per second

at an angle rin~ l\ with the horizon; find the range on the horizontal

plane, the time offlight, and the greatest )ieight attained.

The initial horizontal velocity

= 1280 cos (sin-11) = 1280 .^= 160^63 feet per second,

o

and the initial vertical velocity= 1280x^ = 160 feet per second.

L. D. 12

178 EXAMPLES.

Let T=time to the highest point of the path. Then T is the time in

which a particle starting with a velocity of 160 feet per second, and

moving with acceleration -g, comes to rest.

.'. T= =5 seconds.9

.: greatest height attained

= space described in 5 seconds by a particle starting with velocity 160

feet per second and moving with acceleration - g

= 160 T - ^r 2=160 . 5 - 4 . 32 . 52= 400 feet.

Also the range = distance described in 10 seconds by a particle moving

uniformly with a velocity of 160 ^/63 feet per second

= 1600^/63= 12700 feet approximately.

2. A man throws a stone with a velocity of 64 feet per second to hit a

small object on the top of a wall whose height is 19 feet and whose hori-

zontal distance from the man is 48 feet ; find the direction of projection and

the velocity and direction of the stone when it hits the object.

Let 9 be the angle of projection and T the time that elapses before

the object is struck.

By considering separately the horizontal and vertical motion we have

48 = 64 cos 6 . T, and 19=64 sin . T-

.: eliminating T,

19 = 48 tan 6 - ty =48 tan 5- 9 (l + tan26>).

Hence tan = % or J*-> giving the two possible directions of projection.

Also the vertical and horizontal components of the velocity of the

stone when it hits the object are 64 sin 6 - gT, i.e. 64 sin 0-24 sec 6,

and 64 cos 6 respectively.

Taking the first value of 6 these are respectively f-fv/13 andJT932-^/13 feet per second, whence may be obtained the required velocity

and direction of the stone. Similarly taking the second value of the

velocity corresponding to the alternative path may be found.

3. From the top of a cliff, 80 feet high, a stone is thrown so that it

starts ivith a velocity of 128 feet per second at an angle of 30 with the

horizontal; find where it hits the ground at the bottom of the cliff.

The initial vertical velocity is 128 sin 30 or 64 feet per second, and

the initial horizontal velocity is 64 x/3 feet per second.

EXAMPLES. 179

Let T be the time that elapses before the stone hjts the ground.

Then T ia the time in which a stone projected with velocity 64 and

moving with acceleration - g describes a distance - 80 feet.

.-.- 80= 64 T - i^T

8, and therefore T= 5.

During this time the horizontal velocity remains constant, and hence

the distance of the point where the stone hits the ground from the foot

of the cliff is 320 ^3 feet.

4. A shot leaves a gun at the rate of 500 feet per second ; calculate

the greatest distance to which it could be projected and also the height

it would rise.

Ans. 7812J feet; 1953$ feet.

5. Find the velocity and direction of projection of a shot which

passes in a horizontal direction just over the top of a wall which is 50

yards off and 75 feet high.

Ans. 40^/6 feet per second at an inclination of 45 to the horizon.

6. A projectile fired horizontally from a height of 9 feet from the

ground reaches the ground at a horizontal distance of 1000 feet. Whatis its initial velocity ?

Ans. 1333$ feet per second.

7. A particle is projected horizontally from the top of a tower, 100

feet high, and the focus of the parabola which it describes is in the

horizontal plane through the foot of the tower; find the velocity of

projection.

Ans. 80 feet per second.

8. A particle is projected with velocity 2 *Jog so that it just clears

two walls, of equal height a, which are at a distance 2a from each other.

Shew that the latus-rectum of the path is 2a and that the time of

passing between the walls is 2 fV99. A shot is fired from a gun on the top of a cliff, 400 feet high, with

a velocity of 768 feet per second at an elevation of 30. Find the hori-

zontal distance from the vertical line through the gun of the point where

the shot strikes the water.

Ans. 3200^/3 yards.

122

180 EXAMPLES.

10. Shew that four times the square of the number of seconds in the

time of flight in the range on a horizontal plane equals the height in

feet of the highest point of the trajectory.

11. Find the angle of projection when the range on the horizontal

plane through the point of projection is equal to the height to which the

velocity of projection is due.

Ans. 15 or 75.

12. If the maximum height of a projectile above a horizontal plane

passing through the point of projection be h and a be the angle of pro-

jection, find the interval between the instants at which the height of the

projectile is h sin2 a.

Ans. 2 /27i

V7'13. In a trajectory find the time that elapses before the particle is

at the end of the latus-rectum.

Ans. -(sin acos a).

14. Find the direction in which a rifle must be pointed so that the

bullet may strike a body let fall from a balloon at the instant of firing ;

find also the point where the bullet meets the body, supposing the balloon

to be 220 yards high, the angle of its elevation from the position of the

rifleman to be 30, and the velocity of projection of the bullet to be two

miles per second.

Ans. The rifle must be pointed at the balloon;the bullet will strike

the body when it is 3 inches below the point where it left the balloon.

15. A stone is thrown in such a manner that it would just hit a bird

at the top of a tree and afterwards reach a height double that of the tree;

if, at the moment of throwing the stone, the bird fly away horizontally,

shew that, notwithstanding this, the stone will hit the bird if its hori-

zontal velocity be to that of the bird as ^/2 + 1 : 2.

16. If t be the time in which a projectile reaches a point P of its

path and t' be the time from P till it strikes the horizontal plane through

the point of projection, shew that the height of P above the plane is %gtt'.

17. If at any point of a parabolic path the velocity be u and the in-

clination to the horizon be 6, shew that the particle is moving at right

angles to this direction after a time :-

.

g sm 6

RANGE ON AN INCLINED PLANE. 181

18. In any trajectory shew that the component of the velocity per-

pendicular to the focal radius vector at any point is constant.

19. A particle is projected so as to enter in the direction of its length

a small straight tube of small bore fixed at an angle of 45 to the horizon

and to pass out at the other end of the tube ; shew that the latera recta

of the paths which the particle describes before entering and leaving the

tube differ by ,^2 times the length of the tube.

132. Range on an inclined plane. From a pointon a plane, which is inclined at an angle ft to the horizon,

a particle is projected with a velocity u at an angle a. with

the horizontal in a plane passing through the normal to the

inclined plane and the line of greatest slope, to find the

range on the inclined plane.

Let PQ be the range on the inclined plane, PT the

direction of projection, and QN the perpendicular on the

horizontal plane through P.

The initial component of the velocity perpendicular to

PQ is u sin (a ft), and the acceleration in this direction

is g cos ft.

Let T be the time which the particle takes to go from

P to Q. Then in time T the space described in a direc-

tion perpendicular to PQ is zero.

182 MAXIMUM RANGE.

Hence = u sin (a ft) . T \g cos /3 . T*, and therefore

g cos /3

During this time the horizontal velocity u cos a re-

mains unaltered;hence PN= u cos a. . T.

~ PN u cos a 2w2cos a sin (a /3)

/. the range PQ = ----= = -- =- . T = -2V

cos/3 cos/3 # cosz

/3

133. Maximum range. To find the direction of

projection which gives the maximum range on the inclined

plane, and to shew that for any given range there are two

directions of projection which are equally inclined to the

direction for maximum range.

From the preceding article the range

2w2cos a sin (a

-/9) u* , . ,, m . , ...= - ^ -! = ---

,-Ism (2a

-/3)

- sin /3 . . . (i).

^ cos /3 ^rcos

2

y3l

Now u and are given ;hence the range is a maxi-

7Tmum when sin (2a /3) is greatest or when 2>( yS

= -.

2

7TIn this case a /3 = a, and therefore the required

2

direction of projection, PT, bisects the angle between the

vertical, PL, and the inclined plane.

Hence the maximum range

u* . u*\ ( sin /j )

v

When the range is given let 6 be the required angle

of projection. Then, by (i), sin (20 /9) is given. Nowthere are two values of 20 /3, each less than 180, for

each of which the sine has the same magnitude, and

MOTION UPON AN INCLINED PLANE. 183

these two values are supplementary. Hence if 0,, 2 ,be

the corresponding values of we have

sin (20,-

/3)= sin (202

-/3).

But (^ + /?)has been shewn to be the elevation for

'2i ]

the maximum range.

Hence the two directions of projection for a given

range are equally inclined to the direction of maximum

range.

134. Motion upon an inclined plane. A particle

moves upon a smooth plane which is inclined at an angle

y3 to the horizon, being projected from a point in the plane

with velocity u in a direction inclined at an angle a to the

intersection of the inclined plane with a horizontal plane ;

to find the motion.

Resolve the acceleration due to gravity into two

components ; one, g sin /8, in the direction of the line of

greatest slope and the other, g cos /3, perpendicular to

the inclined plane. The latter acceleration is destroyed

by the reaction of the plane.

The particle therefore moves on the inclined plane

with an acceleration g sin ft parallel to the line of

greatest slope.

Hence the investigation of the motion is the same as

that in Arts. 122 131, if we substitute "^rsin/3" for

"g" and instead of "vertical distances" read "distances

measured on the inclined plane parallel to the line of

greatest slope".

184 EXAMPLES.

EXAMPLES.

1. A plane is inclined at 30 to the horizon ; from its foot a particle

is projected with a velocity of 600 feet per second in a direction inclined

at an angle of 60 to the horizon ; find the range on the inclined plane

and the time of flight.

/3Ans. 2500 yards ; 25^ seconds.

m

2. The greatest range of a particle, projected with a certain velocity,

6n a horizontal plane is 5000 yards ; find its greatest range on an

inclined plane whose inclination is 45.

Ans. 2927 yards approximately.

3. A hill is inclined at an angle of 30 to the horizon ; from a point

on the hill one projectile is projected up the hill and another down ;

the angle of projection in each case is 45 with the horizon ; shew that

the range of one projectile is nearly 3f that of the other.

4. A particle is projected from a point on an inclined plane in

a direction making an angle of 60 with the horizon ;if the range on the

plane be equal to the distance through which another particle would

fall from rest during the time of flight of the first particle, find the

inclination of the plane to the horizon.

Ans. 30.

5. From a point in a given inclined plane two bodies are projected

with the same velocity in the same vertical plane at right angles to one

another ; shew that the difference of the ranges is constant.

6. The angular elevation of an enemy's position on a hill h feet

high isft ;

shew that in order to shell it the initial velocity of the

projectile must not be less than ijgh (l + cosec/3).

7. Shew that the greatest range on an inclined plane through the

point of projection is equal to the distance through which the particle

could fall freely during its time of flight.

GEOMETRICAL CONSTRUCTION. 185

135. Geometrical construction for the path of a par-

ticle projected with a given velocity so as to have a given

range on an inclined plane passing through the point of

projection.

Let P be the point of projection, and u the velocity of2

projection. Draw PK vertical and equal to -

g-

;then

*&KK', the horizontal line through K, is the directrix of

all the paths [Art. 123] and the foci of all the paths lie

on a circle whose centre is P and radius PK. Let this

circle meet PU in H.

To find the focus of the path corresponding to any

range PR we must find a point S on this circle such that

RS may be equal to the perpendicular RK' drawn to the

directrix. A circle whose centre is R and distance RK'cuts the circle of foci in two points S, S' on opposite sides

of PR. Hence there are two paths and the correspond-

ing directions of projection from P bisect the angles KPS,KPS'.

It is plain that the farther R is from P, i.e. the

186 ENVELOPE OF PATHS.

greater the range, the more nearly do S, S' approach, andthe range will be a maximum when S and S' coincide at

H.

Hence the direction of projection for the maximum

range bisects the angle KPU which is the angle between

the vertical and the inclined plane ;also the focus of the

trajectory of maximum range on an inclined plane lies in

the plane.

Again since PS, PS' are equally inclined to PH it

follows that the bisectors of the angles KPS, KPS' are

equally inclined to the bisector of the angle KPH.

Hence the two directions of projection for any given

range on the inclined plane are equally inclined to the

direction for maximum range.

136. Envelope of paths. To shew that the envelope

of all the paths described by particles projected from a

given point with the same velocity is another parabola.

L L'

K'

Let P be the point of projection ;h the height to

which the velocity of projection is due. Draw PK

ENVELOPE OF PATHS. 187

vertical and equal to h. Then the horizontal line KK'is the common directrix of all the paths.

Let PAQ be any one of the paths and let PSQ be its

focal chord through P.

Produce PK to L making KL equal to PK or h and

draw LL' horizontal.

Draw QK'L' a vertical line as in the figure.

Then PQ = PS + SQ = PK + QK1 = K'L' + QK' = QL'.

.'. Q always lies on a parabola whose focus is at Pand of which LL is the directrix and therefore K the

vertex. Let this parabola be the dotted curve.

Since the line QSP passes through the foci of both

parabolas, the tangent to each parabola at Q is the same.

[Art. 120 (5).]

Hence the dotted parabola touches the original para-bola at Q. Similarly it touches all the other paths.

Hence it is their envelope.

The envelope of all the paths is therefore a parabolawhose focus is at the point of projection and whose latus-

rectum is 4h, where h is the height to which the velocity of

projection is due.

1 137. The envelope of the paths can be easily found by analytical

methods. .

For the equation of the path is, by Art. 131,

x2

y = xt&na- r . since u?=2gh,4/i cos-5 o

or y= x tana- (1 + tan2a),

T2 2

or JT tan2 a -* tan a + Tf +y=0.

188 MAXIMUM RANGE.

The equation to the envelope of this curve is (C. Smith's Conic

Sections, Art. 237),

g?= 4.?*L(

This is a parabola whose focus is the point P, whose vertex is K,whose latus-rectum is 4fe, and the concavity of which is turned down-

wards.

138. Since the enveloping parabola touches externally

all the paths, it follows that the maximum range in any

given direction is obtained by finding where this direction

meets the enveloping parabola ;for no point outside the

enveloping parabola can be reached by a particle starting

with the given velocity of projection.

Ex. 1. To find the maximum range on a plane through the point

ofprojection inclined at an angle a to the horizon.

Taking the figure of Art. 136, let PQ, the given inclined plane, meet

the enveloping parabola in Q. Let QM be the perpendicular on PK.

Then 2h = QL' +PM=PQ + PQ , sin a.

.'. the greatest range = PQ = ;

- =7= r .

1 + sin a 0(1 + sin a)

Ex. 2. To find the maximum range on a plane inclined at an angle a

to the horizon and such that the perpendicular distance of the point of

projection from it is d.

Let the plane meet the enveloping parabola in R ; draw EN perpen-

dicular to PK and PY perpendicular to the inclined plane. Then EYis the maximum range required. Let it be x.

Then we have EN 2= h . KN,

.: (x cos a + d sin a)2= 4ft (h + d cos a - x sin a).

Solving this equation we obtain

_ 2 >Jh (h + d cos a)- sin a (2h + d cos a)

cos2 a

which is the required maximum range.



Ex. 1. A particle is projected at an angle a. with the horizontal fromthe foot of a plane, whose inclination to the horizon is ft; shew that it will

strike the plane at right angles if cot ft= 2 tan (a

-ft).

Let it be the velocity of projection so that u cos (a-

ft) and u sin (a-

ft)

are the initial velocities respectively parallel and perpendicular to the

inclined plane.

The accelerations in these two directions are - g sin ft and -g cos

ft.

Then, as in Art. 132, the time, T, that elapses before the particle

. . 2usin(a-/3)reaches the plane again is ~^

If the direction of motion at the instant when the particle hits the

plane be perpendicular to the plane, then the velocity at that instant

parallel to the plane is zero.

Hence u cos (a-

13)- g sin f)T = 0.

u cos (a-

/3) _ T_2u sin (a-

ft)

g Bin ft g cos ft

.: cot]3=2tan(a -ft).

Ex. 2. Particles slide from rest down smooth chords of a vertical

circle, starting from its highest point, and then move freely; shew that

the locus of their foci is a circle of half the size of the given circle

and that their paths bisect the vertical radius.


Let A be the highest point of the circle, its centre, and AP anychord; draw PK perpendicular to AK. Then since (Art. 123) the

velocity at P is that due to the vertical depth of P below A, it follows

that AK is the directrix of the path described by the particle after

leaving P.

Also L OPA = L OAP = L APK, and AP is the tangent at P to the

parabola described.

.. the focus S lies on PO.

Also PS=PK. Therefore the two triangles APK, APS are equal.

Hence ASP is a right angle, and therefore S lies on a circle on AO as

diameter.

Again if B be the middle point of AO, it is the centre of this

circle; therefore SB = BA, and B is therefore a point on the parabola a

portion of which is described by the particle after leaving P.

Ex. 3. A shot of m pounds is fired from a gun of M pounds which is

placed on a smooth horizontal plane and elevated at an angle a. Shew

that, if the muzzle velocity of the shot be V, the range will be

tan'a

Let u be the velocity communicated to the shot along the barrel of

the gun by the explosion of the gunpowder, so that the impulse I of the

force exerted by the expanding gases is mu. An equal and opposite

impulse is communicated to the gun ; let this impulse be resolved into

two, I cos a horizontally, and I sin a vertically.

The latter is counterbalanced by the increased pressure on the

horizontal plane.

Now the shot, whilst inside the barrel, has, in addition to its

velocity u in the direction of the barrel, the same horizontal velocity, U,that the gun possesses, so that we have

(M+m) U=Icosa= mucosa (1).

If F be the resultant velocity, and p the elevation, of the shot as

it leaves the barrel, we have

Fcos j8=itcosa- U, and Fsin/3=wsin a.

EXAMPLES ON CHAPTER V. 191

tan S= '~r,= (1 + ,, )

tan a, by substituting for U from (1).u cos a - U \ MJ

2F28in/3cos/3 2F 2

tan/3TCLTK/A r - v^ _ .

*

g'

o l + tatf

EXAMPLES. CHAPTER V.

1. If v,

v.2 be the velocities of a projectile at the ends of

a focal chord of its path and V the horizontal velocity, shew

that

2. A particle is projected under gravity with velocity u

at an angle a with the vertical;shew that the length of the

at"

focal chord through the point of projection is^ ; 5--

.

3. Shew that the velocity at any point of a trajectory is

equal to that acquired by a body in falling freely through a

vertical distance equal to one quarter of the focal chord

parallel to the tangent at the point.

4. If a be the angle between the tangents at the

extremities of any arc of a parabolic path, v, v' the velocities

at these extremities and u the velocity at the vertex of the

path, shew that the time of describing the arc is -.

5. Shew that the velocity at any point of the path of a

projectile is proportional to the length of the normal at that

point ;hence shew that the velocity is compounded of two

equal velocities, one in a horizontal direction, and the other in

a direction perpendicular to the focal distance of the point.

192 EXAMPLES ON CHAPTER V.

6. On the moon there seems to be no atmosphere and

gravity there is about one sixth of that on the earth. What

space of country would be commanded by the guns of a lunar

fort able to project shot with a velocity of 1600 feet per

second 1

7. Shew that the angular velocity of a projectile about

the focus of its path varies inversely as the distance of the

projectile from the focus.

8. Find the charge of powder required to send a 68 Ib.

shot, with an elevation of 1 5, to a range of 3000 yards, given

that the velocity communicated to the same shot by a charge

of lOlbs. is 1600 feet per second.

9. A coach-wheel rolling with given velocity throws

off small portions of dust at a tangent to its circumference;

find the greatest height from the ground to which any dust

will rise.

10. The radii of the front and hind wheels of a carriage

are a and b, and c is the distance between the axle-trees;

a particle of dust driven from the highest point of the hind-

wheel is observed to alight on the highest point of the front

wheel. Shew that the velocity of the carriage is

(c + b -a) (c + a b)

1

4(6-o)

11. Shew that a speck of mud thrown from the top of a

hansom cab-wheel of diameter d feet moving with a velocity

of v feet per second will, when it strikes the ground, be at

a distance v Jdjg in front of the position then occupied by the

point of contact of the wheel and ground.

Shew also that the mud will not clear the wheel unless v be


12. Three bodies are projected simultaneously from the

same point in the same vertical plane, one vertically, another

at an elevation of 30, and the third horizontally ;if their

velocities be in the ratio 1:1: <J3, shew that they are alwaysin a straight line.

How does this straight line move in space ?

13. Two particles are projected simultaneously, one with

velocity V up a smooth plane inclined at an angle of 30 to the

horizon, and the other with a velocity -^ at an elevation of

60. Shew that the particles will be relatively at rest at the

2Vend of ^ seconds from the instant of projection.

14. A particle, projected with velocity u, strikes a't right

angles a plane through the point of projection inclined at an

angle ft to the horizon. Shew that the height of the pointstruck above the horizontal plane through the point of projec-

2w* sin2 B -tion is -

T ^ Q p^2 Q >that the time of flight is

y

and that the range on a horizontal plane through the point of

u2 sin 2/8 1 -f sin2ft

projection would be - -,

- . .

'

.

g 1 + 3 sin2

ft

15. Two inclined planes intersect in a horizontal line,

their inclinations to the horizon being a and ft ;if a particle

be projected from a point in the former at right angles to it so

that it strikes the latter plane at right angles, shew that its

velocity of projection is sin ft / *9L whereV sin a - sin ft cos (a + ft)

'

a is the distance of the point of projection from the intersec-

tion of the planes.

L. D. 13


16. Two inclined planes of equal height h and inclination

a are placed back to back. A ball projected along the surface

of the plane at an angle ft with the horizontal flies over the

top and falls at the foot of the other plane ;shew that the

velocity of projection is

2 Jgh cosec B JS + cosec2a.

17. A particle is projected from a point in the lowest

line of a plane which is inclined at an angle a to the horizon

with velocity V at an angle (3 to the plane, so that the plane

containing the direction of V and the normal to the plane cuts

the inclined plane in a line making an angle y with the line of

greatest slope. Shew that, if the inclined plane be smooth

and inelastic, the particle will describe two parabolas before it

again reaches the lowest line of the plane and that the time in

this second parabola is

4F ( y y)-. pj- -(cos (a + B) cos2 ~ cos (a

- B) sin2^> .

g sin 2a ( 2 2j

18. A particle is projected so as to just graze the four

upper corners of a regular hexagon, whose side is a, placed

vertically with one side on a table. Shew that the range on

the table is a^/7 and that the square of the time of flight is

19. If the velocity with which a bullet issues from a

rifle be equal to that due to a fall from a height h, shew that,

if the rifle be pointed directly at a point at a height 2A above

a target, a small alteration in the elevation will not affect the

position of the place where the bullet strikes the target.

20. The barrel of a rifle sighted to hit the centre of the

bull's eye, which is at the same height as the muzzle and

distant a yards from it, would be inclined at an angle a to the


horizon. Shew that, if the rifle be wrongly sighted so that

the elevation is (a + 0), where is small compared with a, the

target will be hit at a height'

above the centre of thecos'a

bull's eye.

If the range be 960 yards, the time of flight 2 seconds, and

the error of elevation 1", the height above the centre of the

bull's eye at which the target will be hit is nearly th of an

inch.

21. If a rifle be sighted so as to throw a ball at a small

elevation 8 above the line of aim and thus hit an object distant

R on a horizontal plane, shew that, in order that the particle

may have the same range when firing up or down a given

slope whose inclination is i, the sights must be altered so as to

give an elevation 8 cos i approximately.

22. If a be the angle of projection in order that a bullet

projected with velocity V from a platform at rest may strike

sin object in the same horizontal plane, shew that when

the platform is moving towards the object with a velocity

u (which is small compared with F) the angle of projection

must be diminished by^ s nearly, provided that the' Kcos2a TT

object is well within range for the given velocity.

23. From a fort a buoy was observed at a depression

i below the horizon;a gun was fired at it at an elevation a

but the shot was observed to strike the water at a point whose

depression was i'. Shew that to strike the buoy the gun must

be fired at an elevation where

cos 6 sin (0 + i) cos2i sin i'

cos a sin (a + i')cos2

i' sin i'

24. Shew that the locus of the foci of all trajectories

which pass through two given points is a hyperbola.

132

EXAMPLES ON CHAPTER V.

25. Particles are projected from the same point in

different directions in the same vertical plane so that (1) the

horizontal component of their velocities, (2) the time of flight,

is constant. Shew that the locus of the focus is, in each case,

a parabola.

26. A heavy particle slides down a chord of a vertical

circle to the lowest point and then flies off and describes a

parabola. Shew that the focus of the parabola lies in the

tangent to the circle at the upper extremity of the chord.

27. Particles descend under gravity along various straight

lines from a fixed point to a fixed vertical straight line and

then freely describe parabolic trajectories ;shew that the locus.

of the foci of the paths described is a circle and the locus,

of their vertices an ellipse.

28. A gun on a smootli horizontal plane throws a ball

whose mass is -tli of its own mass: shew that, if then

momentum of the ball on first leaving the gun be constant,

the elevation of the gun giving the greatest range on thefv\

horizontal plane through the muzzle is tan" 1n

-

1 + n

29. A shot is fired with velocity <j2gk from the top of a

mountain which is in the form of a hemisphere of radius r.

Shew that the furthest points of the mountain which can be

reached by the shot are at a distance (measured in a straight

line), r-a/r2 4rA from the point of projection.

30. With given velocity of projection give a geometrical

construction to obtain the maximum range of a ship which

can be struck from a fort on the top of a cliff, and the

maximum range at which the ship can strike the fort;deter-

mine also the angle of projection in each case.


31. Shew that the whole area commanded by a gun

planted on a hill side, supposed plane, is an ellipse whose

focus is at the gun, eccentricity the sine of the inclination of

the hill, and semi-latus-rectum equal to twice the heightto which the velocity of the shot at the muzzle is due.

3*2. A gun is placed on a fort situated on the side of

ft hill, which is a plane of inclination a. Shew that the area

commanded by it is 4irh (h + dcosa) sec3a, where Jlgh is the

muzzle velocity of the shot and d the perpendicular distance of

the gun from the lull side.

33. Shew that the area covered on a vertical wall at

;i distance a by a jet of a fire-engine, placed on level ground,

is, for all directions of the jet, a parabola whose height is

b*-a?(), and breadth 2x/6'

2 -a2,where 6 denotes the maximum

range of the jet.

34. Find that point in a given horizontal plane from

which a particle can be projected with the least velocity so

that it passes through two given points not in the horizontal

plane.

35. Shew that, in order to project a particle with given

velocity v so that it may give the maximum impulse to a

smooth vertical wall at a height h above the point of projection,

yoA

(u2

2gh) from the

vortical wall.

36. The ridge of a roof is 15 feet vertically above the

eaves and the slope is 45;

find the greatest vertical depthbelow the eaves at which a gutter projecting six inches beyondthe eaves may be placed so as to catch all the rain water


coming from the roof, assuming that each drop starts from

rest at the point where it strikes the roof, and neglecting all

friction.

37. A man standing on the edge of a cliff throws a stone

with given velocity u at a given inclination to the horizon, in a

plane perpendicular to the edge of the cliff; after an interval

T he throws from the same spot another stone with given

velocity v at an angle- + with the line of discharge of thefj

first stone and in the same plane. Find T so that the stones

may strike one another, and shew that the maximumvalue of T for different values of 6 is 2v'

2

/gw and occurs when

6 = sin" 1

v/u, where w is v's vertical component.

38. A cannon, whose elevation is a, points at a target ;

the trunnions of the gun are then tilted through an angle ft in

the vertical plane of the axis. Shew that, if R be the range,

the corresponding horizontal deviation is R tan a sinft.

39. Find the charge of powder required with an elevation

of 15 to send a 32 Ib. shot to a range of 1600 yards, it beingknown that the initial velocity of the shot is 1600 feet per

second when the charge is half the weight of the shot.

If the gun be moveable on a smooth horizontal stand, and

if the weight of the gun be n times the weight of the shot,

and the charge that just found, then the range is .

yards.

40. Two parallel lines in the same vertical plane are each

inclined to the horizon at an angle a. From a point midwaybetween them a particle is projected so as to graze one line

and fall perpendicularly on the other;shew that the angle

that the direction of projection makes with either is

tan- 1

[(x/2-l)cota].


41. A particle is projected from the lowest point of a

hollow sphere in such a direction and with such a velocity that

it strikes the sphere at right angles at some point P. If a,

be the angles which the direction of projection and the line

OP respectively make with the horizontal, then

2 tan a = cot 6 + 3 tan 6.

42. Two particles of unequal mass are projected so as to

describe the same path in opposite directions and when theymeet they coalesce

;find the latus-rectum and focus of the

new parabola that they will describe.

43. A bullet is fired in a direction towards a second

equal bullet which is let fall at the same instant. Shew that

they meet and that, if they coalesce, the latus-rectum of the

joint path is one quarter the latus-rectum of the original pathof the first bullet.

44. A thin board, of thickness a, is placed in front of a

gun and opposes a constant resistance nmg to the motion of a

bullet, of mass m, during its passage through the board. The

barrel of the gun makes an angle with the horizon, and the

bullets are tired witli a velocity due to a fall through a

vertical height h;when the board is withdrawn the bullets

hit the top of a tower, and when it is present they hit the

bottom of the tower. Shew that the height of the tower

is -T tan2 6 (h cos 6 no), where n is very large and na finite.

45. A snowball is projected with velocity V horizontally

from the edge of a cliff, a feet in height, and passes through a

layer of cloud, 6 feet in thickness, which is moving horizontally

with velocity parallel and equal to the initial velocity of the

ball;shew that if the mass of the snowball during its passage

through the cloud increase in the time r in the ratio !+-:!,


where u is the vertical velocity of the ball when it reaches the

cloud, the range on a horizontal plane passing through the

foot of the cliff will be

ug

46. The elevation of a rifle for a parabolic trajectory

of range a is a;shew that, taking into account a uniform

horizontal resistance equal to XF2,where V is the initial

velocity, the elevation would have to be increased by

^aX tan a sec 2a

approximately.

47. A particle is projected from a platform with velocity

V at an elevation (3. On it is a telescope fixed at an elevation

a. The platform moves horizontally in the plane of motion of

the particle so as to keep the particle always in the centre of

the field of view of the telescope. Shew that the original

velocity of the platform must be F T- and its accelera-sin a

tion g cot a.

48. A telescope on a heavy platform is drawn up a

smooth plane of inclination a to the horizon by a force so

adjusted that the telescope may keep a given projectile alwaysin the field of view. Shew that, if /? be the angle the

telescope makes with the plane and F the initial velocity of

the projectile perpendicular to the axis of the telescope, the

magnitude of the force per unit mass must be g cos a cot /3 and

the initial velocity of the telescope F cosec ft. The motion of

all the bodies is supposed to be in one plane.

49. A body, of mass m, is projected so as to describe

a parabola under the action of gravity. A blow, whose

impulse is mu, is given to it at any point of its path ;shew


that the focus of the new trajectory moves towards the body

through a distance~"

-w, where v is the original velocity of

*9the body.

50. Two equal particles are connected by a fine inelastic

thread;one is placed on a smooth table and the other just

over its edge, the thread being at full stretch at right angles

to the edge of the table;find the velocity of the centre of

inertia of the particles at the instant after the former has left

the table, and shew that the whole interval that elapses from

the beginning of the motion to the instant when the thread

first becomes horizontal varies as the square root of the length

of the thread.

CHAPTER VI.

COLLISION OF ELASTIC BODIES.

139. IF a man allow a glass ball to drop from his

hand upon a marble floor it rebounds to a considerable

height, almost as high as his hand;

if the same ball be

allowed to fall upon a wooden floor it rebounds through a

much smaller distance.

If we allow an ivory billiard ball and a glass ball to

drop from the same height the distances through which

they rebound will be different.

If again we drop a leaden ball upon the same floors

the distances through which it rebounds are much smaller

than in either of the former cases.

Now the velocities of these bodies are the same on

first touching the floor; but, since they rebound through

different heights, their velocities on leaving the floor must

be different.

The property of the bodies which causes these dif-

ferences in their velocities after leaving the floor is called

their Elasticity.

In the present chapter we shall consider some simple

cases of the impact of elastic bodies. We can only discuss

NEWTON'S EXPERIMENTAL LAW. 203

the cases of particles in collision with particles or planesand of smooth homogeneous spheres in collision with

smooth planes or smooth spheres.

140. Def. Two bodies are said to impinge directlywhen the direction of motion of each is along the commonnormal at the point at which they touch.

They are said to impinge obliquely when the direction

of motion of either or both is not along the commonnormal at the point of contact.

The direction of this common normal is called the line

of impact.

In the case of two spheres the common normal is the

line joining their centres.

141. Newton's Experimental Law. Newton

found, by experiment, that if two bodies impinge directlytheir relative velocity after impact is in a constant ratio to

their relative velocity before impact and is in the oppositedirection.

If the bodies impinge obliquely their relative velocityresolved along their common normal after impact is in a

constant ratio to their relative velocity before impactresolved in the same direction and is of opposite sign.

This constant ratio depends on the substances of

which the bodies are made and is independent of the

masses of the bodies. It is generally denoted by e and is

called the Modulus or Coefficient of Elasticity, Restitution,

or Resilience. Either of the two latter terms is better than

the first, which is used in Physics for other meanings.If u, u' be the component velocities of two bodies

before impact along their common normal, and v, v the

204 COLLISION OF ELASTIC BODIES.

component velocities of the bodies in the same direction

after impact, the law states that v v' = e(u u').

The value of e has widely different values for different

bodies;for two glass balls e is '94; for two ivory ones it is

81;for two of cork it is '65

;for two of cast-iron about

66;whilst for two balls of lead it is about '20 and for

two balls, one of lead and the other of iron, the value is

13.

Bodies for which the coefficient of restitution is zero

are said to be "inelastic

";

whilst "perfectly elastic

"

bodies are those for which the coefficient is unity.

Probably there are no bodies in nature coming strictly

under either of these headings ; approximate examples of

the former class are such bodies as putty, whilst probablythe nearest approach to the latter class is made by glassballs.

More careful experiments have shewn that the ratio of the relative

velocities before and after impact is not absolutely constant but decreases

very slightly for very large velocities of approach of the bodies.

142. Motion of two smooth bodies perpendicular to

the line of impact.

When two smooth bodies impinge there is no tangentialaction between them so that the stress between them is

entirely along their common normal. Hence there is no

force perpendicular to this common normal and therefore

no change of velocity in that direction. Hence the com-

ponent velocity of each body perpendicular to the commonnormal is unaltered by the impact.

143. Motion of two bodies along the line of impact.

From Art. 87 it follows that when two bodies impinge

IMPACT OF A SPHERE ON A SMOOTH PLANE. 205

the sum of their momenta along the line of impact is the

same after impact as before.

The two principles enunciated in this and the previous

article, together with Newton's experimental law, are

sufficient to find the change in the motion of particles

and spheres produced by a collision.

We shall now proceed to the discussion of particular

cases.

144. Impact on a fixed plane. A smooth sphere,

or particle, whose mass is m and whose coefficient of resti-

tution is e, impinges obliquely on a fixed plane ; to find the

change in its motion.

Let AB be the fixed plane, C the point at which the

sphere impinges and CN the normal to the plane at G so

that CN passes through the centre, 0, of the sphere.

Let DO, OE be the directions of motion of the centre

of the sphere before and after impact and let the angles-

206 IMPACT OF A SPHERE ON A SMOOTH PLANE.

NOD, NOE be a and 6. Let u, v be the velocities of the

sphere before and after impact as indicated in the figure.

Since the plane is smooth there is no force parallel to

the plane ;hence the velocity of the sphere resolved in a

direction parallel to the plane is unaltered.

.'. v sin 9 = u sin a. (1).

By Newton's experimental law the relative velocity

along the common normal after impact is ( e) times the

relative velocity before impact.

Hence v cos 9 = e( u cos a 0).

/. v cos = eu cos a (2).

From (1) and (2), by squaring and adding, we have

v = u Jsm* a + e* cos2

a,

and, by division, cot 6 = e cot <z.

These two equations give the velocity and direction of

motion after impact.

The impulse of the pressure on the plane is equal and

opposite to the impulse of the pressure on the sphere and

is therefore measured by the change of the momentum of

the sphere perpendicular to the plane.

Hence the impulse of the blow = mu cos a + mv cos

= m(l + e)u cos a.

Cor. 1. If the impact be direct, we have a = 0.

.'. 6 = 0, and v = eu.

Hence the direction of motion of the sphere is reversed

and its velocity reduced in the ratio 1 : e.

COLLISION OF ELASTIC BODIES. 207

Cor. 2. If the coefficient of restitution be unity we

have 6 = a, and v u.

Hence when the plane is perfectly elastic the angle of

reflexion is equal to that of incidence, and the velocity

is unaltered in magnitude.

Cor. 3. If the coefficient of restitution be zero we

have 6 = 0, and v = u sin a.

Hence the sphere after impact with an inelastic plane

slides along the plane with its velocity parallel to the

plane unaltered.

Cor. 4. If a smooth sphere impinge on a fixed curved

surface the investigation of the impact is reduced to

that of this article, the tangent plane to the surface

at the point of contact being substituted for the curved

surface.

145. If a plane be compelled to move with velocity V and overtake a

particle moving with velocity u in the same direction as the plane, the

velocity v of the particle after the impact is determined by Newton's

law. For, by Newton's law, V-v = -e(V-u).

.: v = (l + e)V-eu.

If the particle overtake the plane, in which case u is greater than V,

the same formula gives the velocity of the particle after the impact. Ina

this case, if V be less than u, the velocity of the particle is reversed-* "T

in direction.

The impulse of the force exerted during the collision

=m (v-u) = TO (1 + e) (

V -u).

146. Direct impact of two spheres. A smooth

sphere, of mass m, impinges directly with velocity u on

another smooth sphere, of mass m', moving in the same

208 DIRECT IMPACT OF TWO SPHERES.

direction with velocity u'. If the coefficient of restitution

be e, to find their velocities after the impact.

Let v, v' be the velocities of the two spheres after

impact.

By Newton's experimental law the relative velocity

after impact is ( e) times the relative velocity before

impact./. v v' = e(u u'} (1).

Again the only force acting on the bodies during the

impact is the blow along the line of centres. Hence, byArt. 87, the total momentum in that direction is unaltered.

/. mv + m'v' mu + m'u' (2).

Multiplying (1) by mf, and adding to (2), we have

(m + m') v = (m em') u + mf (1 + e) u'.

Again multiplying (1) by m, and subtracting from (2),

we have

(m + m') v = m (1 + e) u + (m em) u'.

These two equations give the velocities after impact.

OBLIQUE IMPACT OF SPHERES.

Also the impulse of the blow on the ball m= the change produced in its momentum

209

nun

m= m (u v)

=

The impulse of the blow on the other ball is equal and

opposite to this.

If the second sphere be moving in a direction oppositeto that of the first we must change the sign of u'.

Cor. If we put m = m' and e = I, we have

v = u', and v' = u.

Hence if two equal perfectly elastic balls impinge

directly they interchange their velocities.

147. Oblique impact of two spheres. A smooth

sphere, of mass m, impinges with a velocity u obliquely on

a smooth sphere, of mass m', moving with velocity u'. Ifthe directions of motion before impact make angles a, ft

respectively with the line joining the centres of the spheresand if the coefficient of restitution be e, to find the velocities

and directions of motion after impact.

L. D.

210 OBLIQUE IMPACT OF SPHERES.

Let the velocities of the spheres after impact be v, v'

in directions inclined at angles 0, <f>to the line of centres.

Since the spheres are smooth there is no force perpen-dicular to the line joining the centres of the two balls, and

therefore the velocities in that direction are unaltered.

Hence v sin 6 = u sin a (1),

and v' sin(f>= u' sin 8 (2).

For the motion along the line of centres we have byNewton's Law,

v cos v' cos < = e(u cos a. u' cos /S) (3).

Again, the only force acting on the spheres during the

impact is the blow along the line of centres. Hence

(Art. 87) the total momentum in that direction is

unaltered.

/. mv cos 6 + m'v' cos $ = mu cos a + m'u' cos 8 (4).

The equations (1), (2), (3), (4) determine the unknown

quantities v, v, 0, <f>.

Multiply (3) by m', add to (4), and we obtain

(m em'} u cos a + m' (1 -f e) u' cos 80CO80= r -

(5).m + mSo multiplying (3) by m, and subtracting from (4), we

get

m (1 + e) u cos a + (m1

em) u' cos 8tcos6= ^f -

(6).m + m

From (1), (5) by squaring and adding we obtain v2

,and

by division we have tan 6.

Similarly from (2), (6) we obtain v''2 and tan <.

Hence the motion is completely determined.

OBLIQUE IMPACT OF SPHERES. 211

The impulse of the blow on the first ball = the change

produced in its momentum = m (u cos a v cos 6}

mm . .= .(I+e) (u cos a u cos a), on reduction.m+mThe impulse of the blow on the other ball is equal

and opposite to this.

Cor. 1. If a' = we have from equation (2) <f>= 0, and

hence the sphere m' moves along the line of centres. This

follows independently, since the only force on m' is alongthe line of centres.

Cor. 2. If m = m' and e = 1, we have

v cos 6 = u' cos ft, and v' cos<f>= u cos a.

Hence if two equal perfectly elastic spheres impinge

they interchange their velocities in the direction of the

line of centres.

148. The case of the impact of a particle on a smooth fixed plane

may he deduced from that of two smooth spheres.

If in the last article we make m' infinite and u' zero, we have the

case of a fixed plane. For a sphere of infinite mass at rest may be

looked upon as immovable, and therefore, in the limit, as a fixed plane.In this case the equations (2) and (6) vanish and the equations (1)

and (5) become v sin 6= u sin o,

(m - em') u cos oand v cos = Ltm.=K -

m +m

= Lt,

(m \- e}

\m )mm'

= -eu cos a.

These two equations are now the same as those of Art. 144 if weremember that the velocity of the particle after impact is here measured

In the direction opposite to that in Art. 144.

142

212 EXAMPLES.

EXAMPLES.

1. Two smooth balls, one of mass double that of the other, are moving-

with equal velocities in opposite parallel directions and impinge, their

directions of motion at the instant of impact making angles of 30 with the

line of centres. If the coefficient of restitution be, find the velocities and

directions of motion after the impact.

Let the masses of the balls be 2m, vi, and let the velocities after

impact be v, v' respectively at angles 6, <f>to the line of centres.

Since the velocities perpendicular to the line of centres are unaltered,.

and

(1),

(2).

By Newton's law,

vcos0-t>'cos0= -e[MCOs30-(-ttcos30)] = -^- ....... (3).

Since the momentum resolved parallel to the line of centres remains

unaltered,

.-. 2mv cos 6+ mv' cos <f>= 2mu cos 30 - mu cos 30,

(4).

Solving (3) and (4) we have v cos 6- 0, v' cos<j>= u

^-.

EXAMPLES. 213

From these equations and (1) (2) we obtain

=|,t>

=|; 0=30, v'= u.

Hence after impact the larger ball starts off in a direction perpendicular

to the line of centres with half its former velocity, and the smaller ball

moves as if it were a perfectly elastic ball impinging on a fixed plane.

Also the impulse of the blow

= change in the momentum along the line of centres

= 2w cos 30 =mufJ3.

2. An imperfectly elastic particle is projected from the foot of a plane,

which is inclined to the horizon at an angle p, with a velocity u at an

angle a to the plane. If e be the coefficient of restitution, find the time

tliat elapses before the particle has ceased rebounding and the distance

described by it parallel to the plane in this time.

Find also the condition that it may return to the point of projection

after rebounding n times.

The velocity of the particle perpendicular to the plane initially is

sin o, and this also is its velocity in the opposite direction just before

the first impact. Hence (Art. 144) its velocity perpendicular to the plane

immediately after the first impact is eu sin a; after the second impact

it is e'2u sin a, and so for the velocities after the other impacts. These

velocities form a descending geometrical progression.

Also the acceleration perpendicular to the inclined plane is -g cos p.

Hence the times of describing its trajectories are

2u sin a 2eu sin a 2e2u sin a

g cos p'

g cos p'

g cos p'

The sum of these times

2u sin a ., , . , . 2 sin a 1admf.) =

g cos pv

g cos p 1 - e

After this time the particle ceases to rebound.

Also the distance described parallel to the plane in this time

2u sin a 1 1 . , /2usina 1 \ 2

= w cos a . *7 sin p . I

<7 cos p 1 e '2 \ g cos ^3 1

2u2 sin a=g COs^(l-^ [COS(a +^- eC08gC 8^

214 EXAMPLES.

Again the time, r, of describing n + 1 trajectories

2w sin a2 _ 2w sin a 1 - en+l

U ~~gcosp

If the particle return to the point of projection after rebounding n

times the distance described parallel to the plane in time r must be zero.

.'. = u cos O.T - \g sin /3r2

,

2w cos a_ _ 2u sin a 1 - en+l

g sin ft

~ ~g cos /3

1 - e

:. tan a tan ft (I- en+l)= \-e.

3. If a ball overtake a ball of twice its own mass moving with

one-seventh of its velocity and if the coefficient of restitution between

them be f , the first ball will, after striking the second ball, remain at

rest.

4. A ball of mass 2 Ibs. impinges directly on a ball of mass 1 Ib.

which is at rest ; find the coefficient of restitution if the velocity with

which the larger ball impinges be equal to the velocity of the smaller ball

after impact.

Ans. $.

5. A ball of mass m impinges directly upon a ball of mass 7?^ at

rest ; the velocity of m after impact is iths of its velocity before impact

and the modulus of elasticity is $ ; compare (i) the masses of the two

balls, and (ii) the velocities of m and m^ after impact.

A ns. (i) They are as 3 : 1 ; (ii) they are as 1 : 2.

6. If the masses of two balls be as 2 : 1, and their respective

velocities before impact be as 1 : 2 and in opposite directions, and e be ,

shew that each ball will after direct impact move back with $ths of its

original velocity.

7. Two equal marbles, A and ,lie in a smooth horizontal circular

groove at opposite ends of a diameter ;A is projected along the groove

and at the end of time t impinges on E ; shew that a second impact will

2toccur at the end of time .

e

8. A heavy elastic ball drops from the ceiling of a room and after

twice rebounding from the floor reaches a height equal to one-half that

of the room ;shew that its coefficient of restitution is f]\.

EXAMPLES. 215

9. A particle falls from a height It upon a fixed horizontal plane ; if

e be the coefficient of restitution shew that the whole distance described

by the particle before it has finished rebounding is -2 h, and that the

time that elapses is

9 -

10. The masses of five balls at rest in a straight line form a

geometrical progression whose ratio is 2 and their coefficients of

restitution are each f . If the first ball be started towards the second,

shew that the velocity communicated to the fifth is (f)4

.

11. A series of n elastic spheres whose masses are proportional to

1, e, ea , ...... are lying in a straight line on a smooth table; if the first

impinge directly 011 the second with velocity u, find the resulting velocity

of the last, and shew that -the final kinetic energy is - u- (1- e + en

) wherem

m is the mass of the first ball.

12. A ball of given elasticity slides from rest down a smooth

inclined plane, of length /, which is inclined at an angle a to the

horizon and impinges on a fixed smooth horizontal plane at the foot of

the former ;find its range on the horizontal plane.

Ans. 4el cos a sin2 a.

13. A heavy elastic ball falls from a height of n feet and meets a

plane inclined at an angle of 60 to the horizon; find the distance

between the first two points at which it strikes the plane.

Ans. 2^/3 ne (1 + e) feet.

14. A particle is projected along a smooth horizontal plane from a

given point A in it, so that after impinging on an imperfectly elastic

vertical plane it may pass through another given point B of the

horizontal plane ; give a geometrical construction for the direction of

projection.

Ans. Draw BN perpendicular to the vertical plane and produce to

C so that BN=e . CN; the required direction of projection is AC.

15. A sphere of mass m impinges obliquely on a sphere of mass M.Shew that, if m = eM, the directions of motion of the spheres after impactare at right angles.

16. A sphere impinges on a sphere of equal mass which is at rest ;

if the directions of motion after impact be inclined at angles of 30 to

the original direction of motion of the impinging sphere, shew that the

coefficient of restitution is .

216 EXAMPLES.

17. A ball impinges on another equal ball moving with the same

speed in a direction perpendicular to its own, the line joining the

centres of the balls at the instant of impact being perpendicular to the

direction of motion of the second ball; if e be the coefficient of restitution,

shew that the direction of motion of the second ball is turned through

an angle tan- 1.

18. If two imperfectly elastic spheres impinge on one another, find

the condition that the direction and velocity of one ball after impact

may be the same as the direction and velocity of the other before impact.

Ant. The velocities of the spheres before impact perpendicular to

the line of centres must be the same and their masses in the ratio e : 1,

where e is the coefficient of restitution.

19. Two equal smooth elastic spheres moving in opposite parallel

directions impinge on one another ; if the inclination of their directions

of motion to the line of centres be tan"1^, where e is the coefficient of

restitution, shew that their directions of motion will be turned through a

right angle.

20. Two equal balls are in contact on a table; a third equal ball

strikes them simultaneously and remains at rest after the impact ; shew

that the coefficient of restitution is .

21. Two elastic spheres, P and Q, of masses 1 Ib. and 2 Ibs.

respectively, are placed on a smooth horizontal plane. P impinging

directly on Q at rest drives it perpendicularly against a hard vertical

wall from which it rebounds and meets P when the interval between Qand the wall is one-half what it originally was; find the modulus of

elasticity which is the same for the impact between the spheres and the

plane.

An*.

22. An imperfectly elastic particle is projected from a point in a

horizontal plane with velocity u at an elevation a ; if e be the coefficient

of restitution, shew that it ceases to rebound from the plane at the end

2u sin aof time i

-.

g l-e

23. A sphere, A, impinges directly on a sphere, B, and B then

impinges directly on a third sphere, C; if the spheres be imperfectly

elastic, shew that the velocity communicated to C is a maximum when

the mass of B is a mean proportional between the masses of A and C.

LOSS OF KINETIC ENERGY. 217

149. Loss of Kinetic Energy by impact. Two

spheres ofgiven masses moving with given velocities impinge;

to sheiu that there is a loss of kinetic energy and to find

the amount.

I. Let the collision be direct and the notation as in

Art. 146.

Then we have

mv + m'v' = mu + mu (1),

v v' = e (u u'} (2).

To the square of (1) add the square of (2) multiplied

by mm';we then have

(m2 + mm') v

2 + (m2 + mm'} v'

2

= (mu + mu)2 + e

2 mm' (u u'}2

,

or (m + m'} (mv2 + m'v'*}

(mu + m'u')2 + mm' (u u')

2

(1 e2

)mm' (u u')

2

= (m + m') (mu2 + mu'") (1 e

2

)mm' (u u')

2.

1 e2 mm' . ,s 2

. /. kinetic energy after impact = kinetic energy before

1 e2 mm' . ,N2

impact , (u u) .

2 m + m ^

Hence the loss of kinetic energy is

1 e2 mm'

, 2

2 m + m

and this loss does not vanish unless e = l, that is, unless

the balls are perfectly elastic.

218 LOSS OF KINETIC ENERGY.

II. Let the impact be oblique and the notation as in

Art. 147.

Then we have v sin 6 = u sin a ..................... (1),

v' sin(f)= u' sin ft ..................... (2),

mv cos 9 + m'v' cos = mu cos a + m'u' cos /?. . . (3),

v cos 6 v' cos</>= e (u cos OLU' cos /3). . . (4).

To the square of (3) add the square of (4) multiplied

by mm ; we then have

(m* + mm') v2cos

2 6 + (ra'2 + mm') v'

2cos

2<

= (mu cos a + m'u' cos )" + e2raw' (w cos a u' cos

)

2

= (mu cos a + raV cos /9)2 + wra' (% cos i u' cos /3)

2

(1 e2

) mm' (u cos a u' cos /3)2

.

.'., as in I.,

cos2 + mV2

cos2

< = raw2cos

2a. + ^m'u'

2cos* j3

1 - e2

,,

_N ,---^---> (w cos a w, cos p) ............ (5).

2 m + m

Again, adding together the square of equation (1) mul-

tiplied by jr ,the square of equation (2) multiplied by ,

^ 2

and equation (5), we have

%mv* + bm'v1

(u cos a u' cos yS)8

,u ni> T nl

or kinetic energy after impact = kinetic energy beforeI 2 /

impact , (u cos a u' cos /3)2

,

2 ra + rav

and therefore the energy lost is

1 e2

ra?/i' ,

7 (u cos a u cos pjr.2 m + m

ACTION DURING AN IMPACT. 219

Hence in any impact except where the coefficient of

restitution is unity we see that energy is lost.

This missing kinetic energy is converted into mole-

cular energy and chiefly reappears in the shape of heat.

150. Action between two elastic bodies during-

their collision. When two elastic bodies impinge the time

during which the impact lasts may be divided into two

parts, during the first of which the bodies are compress-

ing one another, and during the second of which they are

recovering their shape. That the bodies are compressed

may be shewn experimentally by dropping a billiard ball

upon a floor which has been covered with fine coloured

powder. At the spot where the ball hits the floor the

powder will be found to be removed not from a geometrical

point only but from a small circle;

this shews that at

some instant during the compression the part of the ball

in contact with the floor was a circle;

it follows that the

ball was then deformed and afterwards recovered its shape.

The first portion of the impact lasts until the bodies

are instantaneously moving with the same velocity; forces

then come into play tending to make the bodies recover

their shape. The mutual action between the bodies

during the first portion of the impact is often called" the

force of compression," and that during the second portion" the force of restitution."

It is easy to shew that the ratio of the impulses of

the forces of restitution and compression is equal to the

quantity e which we have defined as the coefficient of

restitution.

For consider the case of one sphere impinging directly

on another as in Art. 146 and use the same notation.

220 IMPACT OF A PARTICLE

Let U be the common velocity of the bodies at the

instant when the compression is finished. Then

m (u U). is the loss of momentum by the first ball,

and m' (U u") is the gain by the second ball.

/. if I be the impulse of the force of compression we

have I = m (u- U) = m' (U - u'),

.-.-+ ,**u-U+U-u'^u-u' ..... ( 1 ).m m

Again the loss of momentum by the first ball duringthe period of restitution is m

(U v), and the gain by the

second ball is m' (v1

U).

.'. if /' be the impulse of the force of restitution

r = m(U-v) = m'(v'- U),

.: + ,= U-v + v'-U = v'-v..... (2).m m

.'. from (1) and (2)~ = *-^-, = e.

I u u

151. When the bodies which come into collision are

rough, rotations are set up in them and the motion is

much more complicated than when they are smooth. Wecannot discuss the general case within the limits of this

book;the only case we shall consider is that of a particle

impinging on a rough plane.

Impact of a particle on a rough plane. A par-ticle impinges on a fixed rough plane whose coefficient of

friction is p ; to find the resulting change in the motion.

ON A ROUGH FIXED PLANE. 221

Let u, v be the velocities before and after impact, the

directions of motion being inclined at angles a, 6 to the

normal to the plane at the point of impact ;let m be the

mass of the particle.

At each instant during the time the impact lasts the

tangential force is p times the normal force.

/. the total tangential impulse is equal to//,

times the

normal impulse.

/. m (u sin a - v sin 6}= pm (u cos a + v cos 6).

.'. v(jj,

cos 6 + sin 6}= u [sin a /* cos a] (1 ).

Also, by Newton's law,

v cos 6= eucosa (2).

From (1), (2) by division

fi cos 6 + sin _ sin a//,cos a

cos 6 e cos a

or e tan 9 = tan ajj, (1 + e) (3).

Also substituting this value of in (2) we obtain the

value of v.


From (3) it follows that the greater //,the smaller is 6

and then by (2) the less is v. Hence the rougher the

plane the more nearly does the ball rise from it perpen-

dicularly and the less is the velocity of the ball after the

collision.

The student might conclude from equation (3) that it

might be possible, for some values of e, p, and a, that the

value of 6 might be negative, and hence that the particle

might rebound so that its motion was on the same side

of the normal after impact as before the impact. This

however is not the case;for friction, being a passive force,

only acts so as to prevent, or destroy, the velocity of the

particle parallel to the plane. Hence if at any instant

during the time that the collision lasts the velocity

parallel to the plane have been destroyed, the force of

friction immediately vanishes; it could not communicate

to the particle a velocity parallel to the plane in the direc-

tion in which it itself acts.

Ex. 1. A particle impinges on a rough fixed plane, whose coefficient

of friction is ^ ,in a direction making an angle of 60 with the normal

v d

to the plane; if the coefficient of restitution be , shew that the angle of

reflexion of the particle is the same as its angle of incidence and that it

loses half its velocity.

Ex. 2. A particle impinges with velocity u on a rough fixed plane

whose coefficient of friction is f ,in a direction making an angle of 45

with the normal to the plane ;if the coefficient of restitution be

, shew

that the particle rebounds at right angles to the plane with velocity . .

MISCELLANEOUS EXAMPLES. o

1. A smooth ring is fixed horizontally on a smooth table and from a

point of the ring a particle is projected along the surface of the table. If. e

be the coefficient of restitution between the ring and the particle, shew that


the latter will after three rebounds return to the point of projection if its

initial direction of projection make an angle tan"1 e* with the normal to

the ring.

Let ABCD be the path of the particle and the centre of the circle.

Let the angles OAB, OBC, OCD, ODA be 0, <j>, ^, x respectively. Then

ABO, SCO, CDO are also 6, <p, \f/ respectively. If the particle return to

A then DAO is x also.

We have cot$=ecot#; cot ^=ecot 0; cotx = ecot^..........(1);

.-. tan <t>= tan 0; tan $ = - tan 0; tan v= -, tan 0.e e*

'

eA

But, since ABCD is a quadrilateral inscribed in a circle,

.-. tan + tan<f> + tan ^ + tan x= tan tan

<f>tan

\f/ + tan & tan <p tan x

+ tan tan\f/tan x+ tan

<f>tan

\f/tan x-

.-. tan l + - + - + tanS 6 \+ -. + -5 + -

1.

e e4 e5 e6J

= tan3 e .-^ fl+ - + -

e3 [_ e e2

It may be noted that, since tan < tan\]/= ta.n x tan 6= ^tan* = 1,

the angles BCD, DAB are right angles.


2. A particle is placed within a straight tube, of length a and of small

section, and the ends of the tube are closed. If the tube be placed on a

xmooth horizontal plane and projected in the direction of its length, find

the distance advanced by the tube between the first and the (n + l)th impact,

the manses of the tube and particle being equal and the coefficient of resti-

tution between them being e.

Let u be the velocity with which the tube is projected.

The relative velocities of the tube and shot after the 1st, 2nd,...

impacts are respectively-

eu, e*u,- e*u .......

Hence the times that elapse between successive impacts are

a a a

eu'

eau'

e3u

.: time between first and (n + l)th impact

Also throughout the motion the velocity of the centre of inertia is^ >

and hence the distance described by it

u a e~* - 1 _ a e~* - 1=2 eu e'1 - 1

~2 l-e

'

If n be even, the particle is at the (n+ l)th impact in contact with the

end of the tube which initially struck it; if n be odd, the particle is in

contact with the other end of the tube and the centre of inertia has

described a distance greater by s than the tube has.p

Hence the space described by the tube is

a e~n - 1 a e~n - 1 a

2T^7 r2 l-e ~2'

according as n is even or odd.

3. A particle is projected from a point in a smooth plane inclined to

the horizon at an angle a, in a vertical plane cutting the inclined plane in

a horizontal line, and in a direction inclined to the horizon at an angle 0.

Shew that at the nth rebound the distance described by the particle paral-

e (1 - en-1 )

lei to the line of greatest slope is a sin a tan v -, ichere a is the

horizontal distance described and e is the coefficient of restitution.


If u be the initial velocity of the particle, its initial horizontal velocity

is u cos 6, and its initial vertical velocity is u sin 0.

The latter is equivalent to two components, u sin 6 cos a perpendicular

to the inclined plane, and u sin 9 sin a parallel to the line of greatest slope.

Also the acceleration in these two directions is g cos a and - g sin a

respectively.

After each impact the velocity perpendicular to the plane is altered in

the ratio of 1 to e.

Hence the time, T, that elapses before the nth impact

2ttsin0cosa r , _ ,.,2u sin l-en~

... .

g cos a g 1 - e

During this time the horizontal velocity remains unaltered,

/. a=u cos . T.

Also if x be the distance described parallel to the line of greatest slope

we have x=u sin 8 sin a . T - \g sin aT 2.

/.-= sin a ["ten-

n-^

a . T\ = sm a tan [~1-\a [_ 2ucos0 J L 1-J

a.-. x= - a sin a tan -

1-

4. A ball is projected from a point A at an elevation a against a

rough vertical wall and in a plane perpendicular to the wall ; shew that

after impact with the wall it will return to the point of projection if

ga (1 + e) cos X= 2eV2 cos a sin (a-

X), where V is the velocity ofprojection,& the distance of A from the wall, e the coefficient of restitution, and tan X

the coefficient offriction between the ball and the wall.

The time, T, in the first trajectory is and the time, 2\, inFcos a

the second is --=;- .

eV cosa.

Also the impulse of the blow normal to the wall is (1 + e) Fcos a and

therefore the tangential impulse is /x (1 + e) V cos a.

Hence if U be the vertical velocity after the impact we have

t*.(l+ e) Vcosa=Vsina-g T- U..................... (1).

Now in the time T+ Tlthe whole distance described vertically is zero.

.-. = 7 Bin a T - %gT*+ VT^ -faT^

2,

L. D. 15


or by (1), substituting for U,

0=Fsino (r+ Tj) -fer (T+ T^-p (1 + e) V cos aTr

.-. 0= Fsin a %g 1 + e 2\- n V cos a, since T=eTv

.-. 2eF2 cos o sin (a-

X) =ga (1 + <) cos X.

5. An inclined plane, of mast M, is capable of moving freely on a

smooth horizontal plane and a smooth sphere, of mass m, is dropped upon

its face and rebounds ; shew that the initial velocity of the plane is

m (1 + e) u sin a cos a / (M + m sin2 a), where u is the velocity of the sphere

on striking the plane, a it the inclination of the face of the plane, and e

is the coefficient of restitution.

Immediately after the impact let Uv J7a be the components of the

velocity of the ball respectively perpendicular to and along the plane, and

let 7 be the velocity of the plane.

.(1).

Since the plane is smooth, we have

C72=u sin a

Also by Newton's Law, we have

- U1- Fsino= -eu cos o (2).

Also the impulse perpendicular to the inclined face=m (wcoso+ UJand the horizontal component of this impulse is equal to the momentum

generated in the plane.

/. m (u cos a + Uj_) sin a=MV

EXAMPLES ON CHAPTER VI. 227

or MV-mU1sin a = mu cos a sin a (3).

From (2), (3), by solving, we have

V(M+msiu?a) =mu (l+e) cos a sin a (4)

and U1 [M+msin2a] = MCOSa (eM - m sin2

a) (5).

Equation (4) gives the velocity of the plane and (1), (5) give the

velocity and direction of the sphere.

EXAMPLES. CHAPTER VI.

1. SHEW that, in order to produce the maximum deviation

in the direction .of a smooth billiard ball of diameter a by

impact on another equal ball at rest, the former must be

projected in a direction making an angle sin"1 -

. / withC v O^6

the line of length c joining the two centres, and prove also

l+ethat this maximum deviation is tan" 1

;

. where e is

2V2\/l-ethe coefficient of restitution.

2. A series of balls, whose masses are M-^ M2i ,are

arranged with their centres in a straight line and the coefficient

of restitution between the rth and (r + 1 )th is ,r.

+l;shewM

r

that, if M1 impinge directly on M2 at rest and so on, the

velocity of each ball between the impacts is equal to the initial

velocity of Mr

3. The red ball is standing in the middle of a billiard

table whose length is 2a and breadth 2b;shew that the

direction in which the white ball must be sent from baulk so

that it may hole the red ball in the top pocket and itself in

the other is inclined to the side of the table at an angle

tan" 1

7-; T5T >and hence find limits to the possibility of

a (a2 eb2)

.the stroke.

152

228 EXAMPLES ON CHAPTER VI.

4. An inelastic ball, of small radius, sliding along a

smooth horizontal plane with a velocity of 16 feet per second

impinges on a smooth horizontal rail at right angles to its

direction of motion;

if the height of the rail above the plane

be one half the radius of the ball, shew that the latus rectum

of the parabola subsequently described is one foot in length.

5. Particles are projected from the same point with equal

velocities; shew that the vertices of their paths lie on an

ellipse, and if they be equally elastic and impinge on a vertical

wall then the vertices of their paths after impact lie on an

ellipse.

6. Two equal smooth balls, whose elasticity is,and

which are in the same horizontal plane at a distance 2a apart

from each other, are projected with the same horizontal velocity

Jga toward one another and are acted on by gravity only ;

shew that, after collision, the velocity of each ball will be

Jfya, and find the position of the vertices and foci of the

subsequent paths.

7. A square is placed in a vertical plane with two sides

vertical and from the foot of one of these a particle is pro-

jected with 2/ times the velocity that would be acquired in

falling down it, the direction of projection being inclined at an

angle cos' 1 i to the horizontal. Find the coefficient of elas-

ticity at the opposite vertical side in order that the particle

after striking the highest side (coefficient of elasticity ^) mayreturn to the point of projection.

8. A perfectly elastic ball is at the focus of an elliptic-

billiard table; shew that the ball however struck will ulti-

mately be moving along the major axis.

9. A ball at the focus of an ellipse, whose eccentricity is

c, receives a blow and after one impact on the elliptic perimeter


passes through the further end of the major axis. Find the

point of impact on the ellipse and shew that c cannot be

greater than .

10. The sides of a rectangular billiard table are of lengths

a and b. If a ball of elasticity e be projected from a point in

one of the sides, of length b, to strike all four sides in

succession and continually retrace its path, shew that the

angle of projection with the side is given by ae cot 6 = c + ec',

where c and c' are the parts into which the side is divided at

the point of projection.

11. Two equal balls, of radius a, are placed in contact on

a rectangular horizontal table bounded by vertical walls so

that the line joining their centres is parallel to an edge of

length h; they are projected simultaneously towards this edge

in directions inclined at angles of 45 with it;shew that, sup-

posing perfect resilience, a series of collisions will occur

between the balls at intervals of time equal to J2 (h-

%a)/v.

12. A smooth circular table is surrounded by a smooth

rim whose interior surface is vertical. Shew that a ball of

elasticity e projected along the table from a point in the rim in

^__ with the radius1 + e + e*

through the point will return to the point of projection after

two impacts on the rim. Prove also that when the ball

returns to the point of projection its velocity is to its original

velocity as e . 1.

13. Two equal elastic balls not in the same vertical line

are dropped upon a hard horizontal plane ;if the balls ever

come simultaneously to a position of instantaneous rest, shew

that the initial heights of the balls above the plane are as

(em-l)

2 to (en-l)

2


where m, n are positive integers and e is the coefficient of

elasticity.

14. A number of particles are let fall from a horizontal

straight line on the convex arc of a fixed vertical parabolawhose directrix is the fixed line; shew that the parabolaswhich they describe after impact on the curve have a commondirectrix at a distance a (1 e

2

)below the given line, where 4a

is the latus rectum of the given parabola and e the coefficient

of restitution between the particles and the curve.

15. A number of perfectly elastic particles are projectedat the same instant from a point in a horizontal plane of

unlimited extent. Shew that the path of the centre of inertia

of these particles will be a curve with a succession of salient

points, the arcs between them being arcs of the same parabola.

16. An imperfectly elastic ball is projected with velocity

Jgh at an angle a with the horizon so that it strikes a

vertical wall distant c from the point of projection and

returns to the point of projection. Shew that the coefficient

cof restitution between the ball and the wall is

h sin 2a - c'

17. A ball of elasticity e is projected from a point in a

smooth vertical wall against a parallel wall, and after 2n + 1

impacts it returns to the point of projection. If the angle of

projection be 60 and the velocity of projection be F, shew

that

where a is the horizontal distance between the two walls.

18. A particle is projected from a point at the foot of one

of two smooth vertical walls so that after three reflexions it

may return to the point of projection and the last impact be


direct; shew that e3

-f e* + e= 1, and that the vertical heights

of the three points of impact are as e2

: 1 - e2

: 1.

19. If two small perfectly elastic bodies be projected at

the same instant with velocities which are as 2 tan ft to

J\ 4- 4 tan2(3, one up a plane inclined at an angle ft to the

horizon, and the other in the same vertical plane but in

a direction making an angle 6 with the plane such that

2 tan 0=cot ft, they will return to the point of projection at

the same instant.

20. A particle of elasticity e is dropped from a vertical

height a upon the highest point of a plane, which is of length

b and is inclined at an angle a to the horizon, and descends to

the bottom in three jumps. Shew that

b = 4a sin ae (1+ e) (1 -f e + 2e2 + e3 + e

4).

21. A particle is projected from the foot of a plane

inclined at an angle ft to the horizon in a direction makingan angle a with the inclined plane and e is the coefficient of

restitution between the particle and the plane. If after

striking the plane n times the particle rebound vertically,

shew that

1 + ecos (a

-ft)

= sin a sin ft .

^ (1- e

n).

Shew also that if 2 tan a tan ft be > 1 - e, the particle will

after a certain number of impacts rebound down the inclined

plane.

22. A particle is projected from the foot of an inclined

plane and returns to the point of projection after several

rebounds, one of which is perpendicular to the inclined plane ;

if it take r more leaps in coming down than in going up, shew

that

cot a cot 6 = {2 Vl-er - 2(1- e")}

+(1-

e) er

,


where a is the inclination of the plane, the angle between

the direction of projection and the plane, and e the coefficient

of restitution.

What is the condition that it may be possible to projectthe particle so that one of its impacts is perpendicular to the

plane ?

23. A particle is projected with velocity F at an angle a

with the horizon and at the highest point of its path impingeson a wall inclined at an angle ft to the horizon

;shew that

after rebounding it will pass over the top of the wall if its

height be

F2

~ {sin2 a cos2 B + 4.e cos2 a sin2 B (cos

2 B - e sin2/? )},

2g cos p

and find the total range on a horizontal plane passing throughthe point of projection.

24. A ball, of which e is the modulus of elasticity, after

dropping through a height h strikes at a point A a planeinclined to the horizon at an angle a and afterwards passes

through a point B in a horizontal line through A;find the

time of moving from A to B, and shew that the problem is

impossible if e be < tan2a.

25. A ball, of elasticity e, is projected obliquely up an

inclined plane so that the point of impact at the third time of

striking the plane is in the same horizontal line with the point

of projection ;shew that the distances from the line of the

points of first and second impact are as 1 : e.

26. A particle, of elasticity e, is projected with a given

velocity so that the whole distance measured up an inclined

plane passing through the point of projection and described

whilst parabolic motion continues is a maximum, the whole

motion taking place in a vertical plane. Shew that if a


be the inclination of the plane and ft the angle which the

direction of projection makes with the inclined plane, then

tan 2ft (1 e) cot a, and that the time that elapses before

the parabolic motion ceases is to the whole time during which

the particle is moviug up the plane as cos 2(3 is to cos2

ft.

27. An imperfectly elastic particle, whose coefficient of

elasticity is e, is projected from a point in a plane inclined to

the horizon at an angle y. The direction of projection makes

an angle ft with the normal to the plane and the plane throughthese two lines makes an angle a with the line of greatest

slope on the inclined plane. When the particle meets the

plane for the nth time it is in the horizontal line through the

point of projection ;shew that (1 e

n)tan y = (1 e) tan ft cos a,

and find the distance of the particle from the point of pro-

jection.

28. A perfectly elastic particle acted on by no forces is

projected from the centre of a square and rebounds from its

sides;

if it ever pass through an angular point, shew that the

direction of projection makes with a side an angle tan" 1

^

where m and n are integers, and if it ever strike the middle

Q*vi _1_ 1

point of a side the angle must be of the form tan" 1 -.

2n

29. Three smooth billiard balls, each of radius d, and with

coefficient of resilience unity, rest on a smooth table, their

centres forming a triangle ABC ;shew that if the ball A is to

cannon off B on to C the angle of impact at B must lie between

TT 2d cos BB - - tanC - 2d sin B

D 9 TT 2d cos (B + 8) .. 4dand B + b - - - tan J

p, .

v 7

,where 8 = sin J

2 (7- 2a sin (B + 8)a


30. A man has to make a small billiard ball, A, 'strike

another ball, B, after one impact 011 a cushion. He knows the

direction in which to strike A but is liable to a small angularerror in striking. Shew that the positions of B in which it is

equally liable to be struck by A lie on a curve whose polar

equation referred to a certain origin can be put into the form

r2

(e2 sina + cos2

6)= const.,

where e is the coefficient of restitution.

31. A perfectly elastic ball is projected horizontally from

the top of a series of steps whose breadths are just greater

than double their heights with a velocity equal to that which

would be acquired in falling through the height of one step ;

shew that it falls on the next step and clears 2, 4, 6, ...steps in

subsequent rebounds.

32. A perfectly elastic ball is thrown into a smooth

cylindrical well from a point in the circumference of the

cylindrical mouth. If the ball be reflected any number of

times from the surface of the cylinder, shew that the times

between the successive reflections are equal. If it be projected7T

horizontally in a direction making an angle- with the tangent7I

at the point of projection it will meet the water at the instant

of the nth reflection provided that the velocity of projection is

nr sin - A /~ ,where r is the radius and d the depth of the

n V d '

well.

33. A hollow elliptic cylinder is placed on a horizontal

table with its axis vertical. From the focus of a horizontal

section a perfectly elastic ball is projected with velocity v in a

horizontal direction. Shew that if the particle return to the

point of projection then the height of the section above the

table is 2m20-

2

/w2 v2

,where m, n are positive integers and 2a

the major axis of the section.


34. Three equal balls A, B, 0, follow one another in this

order in a straight line such that u>v>w. The following

impacts take place in order;A strikes B, B strikes C, and A

strikes B again. If the coefficient of restitution be \ for all

the balls and 4/40 be > 25v- 2lu, shew that no more impacts

will happen.

35. Two equal balls, A and B, are lying very nearly in

contact on a smooth horizontal table and a third equal ball

impinges on A, the centres of the three balls being in the same

straight line;shew that, if e be > 3 - 2^/2, there will be three

impacts, and that the final velocity of B will be to the initial

velocity of the striking ball as (1 + e)2

: 4.

36. Two equal balls, A and B, of elasticity e, move on a

horizontal plane with uniform velocity in the same straight

line which is perpendicular to a given wall and B impinges on

the wall;shew that two impacts between A and B must follow

and two between B and the wall, and that there will be a third

collision between the balls if e be <2 \/3.

37. Two particles, of masses m and m', are constrained to

remain in a smooth circular groove, of radius a, which is cut

in a horizontal table;m is projected along the groove with

velocity u and impinges on m';m' then impinges on m and so

on;shew that the kinetic energy will never be so small as

--;x initial kinetic energy, and that the time betweenm + m

i

the first and nth impacts is - -^--,where e is the

u en 1

(1-e)coefficient of restitution.

38. Two equal balls of radius a are in contact and are

struck simultaneously by a ball of radius c moving in the

direction of their common tangent ;if all the balls be of the

same material and the coefficient of restitution be e, find


the velocities of the balls after impact, and shew that the

impinging ball will be reduced to rest if 2e = -~--.

3

39. A bucket and a counterpoise connected by a string

passing over a smooth pulley just balance one another and an

elastic ball is dropped into the centre of the bucket from a

height h above it;find the time that elapses before the ball

ceases to rebound, and shew that the whole descent of the

bucket during this interval is =-Y>-- n-r, , where m. M are2M+m (1

-ef

the masses of the ball and bucket and e is the coefficient of

restitution.

40. If a portion of a horizontal plane begin to move

upward under the action of its weight W and a constant force

upward, and at the same instant a particle of weight w and

coefficient of restitution e be allowed to drop from a height h

above it, shew that by the time the particle is at rest relative

to the plane the height travelled by the latter is less than the

height through which it would have travelled had no particle

wh 4edropped on it by T

-- --^ .W + w (1 -ef

41. A vertical shower of hail begins to fall with velocity

u and strikes a horizontal slab of unit area. If M be the mass

of the particles in a unit volume of the shower before impact,

2eushew that at time :--r the pressure on the slab is

?(!*)1 +e

MU?^-

,where e is the coefficient of elasticity.

1 Q

What must be taken into account in the measure of the

pressure after this time 1

42. A smooth circular ring, of mass M and radius a,

rests on a smooth horizontal table, and a small spherical ball


of mass m is projected from the centre of the circle with

velocity v ; shew that the whole time that elapses before the

nth impact is -^.-j j ,

where e is the coefficient of

restitution.

Find also the velocity of the ring and sphere after an

infinite time.

43. A boy standing on a bridge lets a ball drop upon the

horizontal roof of a railway carriage passing under him at the

rate of 15 miles per hour. If the coefficient of elasticity

between the ball and the roof be an(i the coefficient of

friction be |, find the least height of the boy's hand so that the

ball may rebound from the same point. If the boy's hand be

at a less height than this, what will happen ?

44. A particle is projected with velocity V from a point

in a horizontal plane in a direction making an angle a with

the plane. The coefficient of elasticity between the particle

and the plane is e and the coefficient of friction tan X. Shew

1 ethat, if tan a be <

^ cot X, the particle will come to rest on1 + e

the plane after a time -. r- and that the distance

g sin X

described measured along the plane from the point of projection

V'2 cos2(a -A)T*

(_

g sin 2X

I -eHow will the problem be modified if tan a be > r- cot X?

1 +e

45. Two equal spheres, A and B, are lying in contact 011

a smooth table. A is struck by a third equal sphere movingin a direction making an angle a with AB

;shew that the

direction of motion of A after the impact is inclined to AB at

an angle tan" 1

(2 tan a).


46. A smooth ball, of mass m', is lying on a horizontal

table in contact with a vertical inelastic wall and is struck byanother ball of mass m moving in a direction normal to the

wall and inclined at an angle a to the common normal at the

point of impact ;shew that, if the coefficient of restitution

between the two balls be e, the angle through which the

direction of motion of the striking ball is turned is given bythe equation

m +m sin2 atan (6 + a) - - tan a

,

-; r- .

em msixra

47. A billiard ball in contact with a cushion is struck

obliquely by another ball, the line joining their centres makingan angle 6 with the cushion

;the coefficient of elasticity for

each impact is e and the moments of greatest compres-sion are simultaneous. Shew that if the striking ball move

parallel to the cushion after impact its velocity then is to that

of the other as 1 - e sec2

: 1 + e.

48. A smooth uniform hemisphere of mass M is sliding

with velocity V on an inelastic plane with which its base is in

contact;a sphere of smaller mass m is dropped vertically and

strikes the hemisphere on the side towards which it is movingso that the line joining their centres makes an angle of 45

with the vertical; shew that, if the hemisphere be stopped

dead, the sphere must have fallen through a height

where e is the coefficient of restitution between the sphere and

hemisphere.

49. A smooth sphere, of mass m, is tied to a fixed point

by an inelastic string and another sphere, of mass m', impinges

directly on it with velocity v in a direction making an acute

angle a with the string. Shew that m begins to move with


m' (1 -f e) sin avelocity

-.

-v, and find the change in the velocitym + m sin-

1 a

of m'.

50. Two inelastic spheres, of masses m and m', are in

contact and m receives a blow through its centre in a direc-

tion making an angle a with the line of centres. Shew that

the kinetic energy generated is less than if m had been free in

the ratio m + m sin2 a to m + m'.

51. Three equal balls of mass m lie in contact on a

smooth table so that their centres are at the corners of an

equilateral triangle ;a ball of mass M impinges directly

with velocity U on one of the balls so that its direction of

motion produced bisects the angle of the triangle ;shew

that the velocity of M after impact is U =-= - and the2M + 5m(1 -f e) M

velocity of the ball struck is 2U \ , r'K , assuming that the2M + 5m

compression ends at the same moment for all the balls.

52. Two equal inelastic balls of masses m,, maare in

contact, and a ball, of mass M, strikes both simultaneously ;

shew that its direction of motion is turned through an angle

-i (mi

~ mz)

sin a cos atan -JT 7-

r ; 5 ,M + (m j+ ra2) sin^a

where 2a is the angle subtended at the centre of M by the

line joining the centres of the other two balls.

53. Three equal balls A, JB, C moving with the same velocity

v in directions inclined at angles of 120 to one another

impinge so that their centres form an equilateral triangle.

If the coefficient of elasticity between C and A or B be e and

between A, B be e', shew that A, B separate with velocity

it being assumed that the compression ends at the same time

for all the balls.


54. Two smooth balls of masses m^, m^ are connected by

an inextensible string and lie at rest at the greatest possible

distance apart. A ball of mass M impinges directly on the

ball m1with a velocity F the direction of which makes an

angle a with the string. Shew that the velocity of M after

the impact is

Jsin2 a cos2 a e

} ^ (sin2 a cos8 a

_1_|

t w*i m^ + m^ M) \ m^ m^ + m2 M)'

where e is the coefficient of restitution of the balls. Find also

the velocity of the ball that is struck.

55. A smooth fixed horizontal tube lies on a smooth table

and half out of it, and just fitting it, lies a smooth ball of mass

A. A second ball of mass B projected along the table in a

direction perpendicular to the tube strikes A so as to drive it

into the tube. Shew that after impact the direction of B'a

motion makes with the common normal an angle

, ( . B cos2 6 + Atan 1

1 cot -=-5-3 T ,

Bcos*6-Ae)

where 6 is the angle between this normal and the axis of the

tube, the centres of both balls being supposed to move in the

same horizontal line.

56. An inelastic particle falls vertically upon a smooth

wedge of angle a which rests on a horizontal smooth table.

Shew that after impact the particle will describe a straight

line with uniform acceleration and that its velocity after

sliding off the wedge on to the table will be

/*/V

sin a cos a /, (rt+l)coseca- ^

* -r .

1 + n V 1 + n sin- a

where V is the velocity with which the particle strikes the

wedge at a distance a from its foot, and n is the ratio of the

mass of the particle to that of the wedge.

CHAPTER VII.

THE HODOGRAPH AND NORMAL ACCELERATIONS.

152. IN the following chapter we shall consider the

motion of a particle which moves in a curve. It will be

convenient, as a preliminary, to explain how the velocity,

direction of motion, and acceleration of a particle movingin any manner may be marked out by means of another

curve.

Hodograph. Def. If a particle be moving in any

path whatever and if from any point O, fixed in space,

we draw a straight line OQ parallel and proportional to the

velocity at any point P of the path, the curve traced out bythe end Q of this straight line is called the hodograph ofthe path of the particle.

153. Theorem. If the hodograph of the path of a

moving point P be drawn, then the velocity of the corre-

sponding point Q in the hodograph represents, in magnitudeand direction, the acceleration of the moving point P in

its path.

L. D. 16

242 THE HODOGRAPH.

Let P, P be two points on the path close to one

another;draw OQ, OQ' parallel to the tangents at P, Pf

and proportional to the velocities there, so that Q, Q' are

two points on the hodograph very close to one another.

Whilst the particle has moved from P to P' its velocity

has changed from OQ to OQ, and therefore, as in Art. 24,

the change of velocity is represented by QQ'.

Now let P' be taken indefinitely close to P so that

QQ' becomes an indefinitely small portion of the arc of the

hodograph.If T be the time of describing the arc PP, then, by

QQ'Art. 26, the acceleration of P = limit of - - =

velocity of QT

in the hodograph.Hence the velocity of Q in the hodograph represents,

in magnitude and direction, the acceleration of P in the

path.

154. Examples. The hodograph of a point describing

a straight line with constant acceleration is a straight line

which the corresponding point describes with constant

velocity.

NORMAL ACCELERATION. 243

The hodograph of a point describing a circle with

uniform speed is another circle which the corresponding

point describes with uniform speed.

The hodograph of the path of a heavy particle pro-

jected freely is a vertical straight line which the corre-

sponding point describes with uniform velocity.

This proposition may be easily shewn without reference to the last

article. For if v be the velocity at any point P of a trajectory we have,

by Art. 123, v"-=2g.SP. But if SY be the perpendicular upon the

tangent at P then SY*=AS . SP so that v*=^-SY*.^1O

.. v is proportional to SY.

Hence if through S we draw a line SY' perpendicular and proportionalto SY the end Y' will describe the hodograph. Also since the locus of Yis the tangent at the vertex of the parabola the locus of Y' will be a

vertical straight line.

Normal Acceleration.

155. We have learnt from the First Law of Motion that

every particle, once in motion and acted on by no forces,

continues to move in a straight line with uniform velocity.Hence it will not describe a curved line unless acted uponby some external force. If it describe a curve with uni-

form speed there can be no force in the direction of the

tangent to its path or otherwise its speed would be altered

and so the only force acting on it is normal to its path.If its speed be not constant there must in addition be a

tangential force.

In the following articles we shall investigate the

normal acceleration of a particle moving in a given

path.

162

244 NORMAL ACCELERATION.

15G. Theorem. If a particle describe a circle ofradius r with uniform speed v, to shew that its accele-

v2

ration is directed toward the centre of the circle.

Let P, P' be two consecutive positions of the moving-

point and Q, Q' the corresponding points on the hodograph.Since the speed of P is constant the point Q moves on a

circle whose radius is v;also the angle QO'Q' is equal to

the angle between the tangents at P, P' and therefore to

the angle POP'.

Hence the arc QQ : the arc PP' :: O'Q : OP :: v : r.

/. the velocity of Q in the hodograph : v :: v : r.

v*.*. velocity of Q .

But the point Q is moving in a direction perpendicular

to O'Q and therefore parallel to PO;also the acceleration

of the point P is equal to the velocity of Q.

v*Hence the acceleration of P is - in the direction PO.

r


Cor. 1. If to be the angular velocity of the particle

about the centre 0, we have v = ra>, and the normal

acceleration is o>V.

Cor. 2. The force required to produce the normal

v*acceleration is m where m is the mass of the particle.

Cor. 3. If the path of the particle be a curve whose

radius of curvature at P is p and its speed be constant the

same proof will apply if we suppose to be the centre of

v2

curvature at P; in this case the acceleration is along

the normal at P.

157. In the present article we shall give a more

general proof of the theorem of the preceding article in

which we shall not suppose the speed in the curve to be

constant, and we shall also find the tangential acceleration.

O'

Let P, P' be two consecutive points in the path of the

particle at which the speeds are v and v' and let the

normals at P, P' meet in 0. Draw O'Q, O'Q' parallel and

proportional to v and v' so that QQ' is an element of the

arc of the hodograph and therefore represents the changeof velocity of the moving point in magnitude and direc-

tion.

246 NORMAL ACCELERATION.

Draw $N perpendicular to QN. Then the velocity

QQ' is equivalent to the velocities QN, NQ' which are

parallel to the tangent and normal at P.

Let T be the time of moving from P to P' and let the

radius of curvature at P be p so that we have

P= PF = vr ultimately, where 6 = POP' = QO'Q'.

T> J/. normal acceleration at P =

T , , T.V'.V v2

, .= Ltv .- = Lt- =- ultimately,T p p

when P and P' are taken indefinitely close to one another.

Also the tangential acceleration at P

QN T ON OQ r v' cos v= Jut = -Lit Jut

T T T

V V= Lt -, ultimately.

T

Hence the tangential acceleration of the particle is

equal to the rate of change of its speed.

158. Without the use of the hodograph a proof of the very important

theorem of Art. 156 can be given as follows.


Let P be a point on the path close to P. Draw the tangent P'T at P1

to meet the tangent, Px, at P in T, and let the normals at P, P* meet

in 0, which, when PP' becomes very small, becomes the centre of

curvature at P.

Since the angles at P, P are right angles a circle will go through

0, P, T, P' and hence PTx - supplement of PTP = POP' = 6.

Let v, v' be respectively the velocities at P and P' and let T be the

time of describing the arc PP'.

Then in time T a velocity parallel to PO has been generated equal to

v' sin 6.

Hence the acceleration in the direction P0= Limit ofT

.Bine 6 arcPP'= Liimt of v . . ^ . .

e arc PP' T

But in the limit when P' is taken indefinitely close to P we have

sin0 .I arcPP'

v =v: ^= -

DD/~

arc PP' p

/. required acceleration = v .-

. v = .

As before, the force in the direction of the normal to cause this

acceleration must be m .

P

159. The force spoken of in the preceding articles

which is required to cause the normal acceleration may be

produced in many ways. For example, the particle may be

tethered by a string, extensible or inextensible, to a fixed

point ;or the force may be caused by the pressure of a

material curve on which the particle moves; or, again,

the force may be of the nature of an attraction such

as exists between the sun and earth and which compelsthe earth to describe a curve about the sun.

160. When a man whirls in a circle a mass tied to

one end of a string, the other end of which is in his hand,

the tension of the string exerts the necessary force on the

248 EXAMPLES.

body to give it the required normal acceleration. But, bythe third law of motion, the string exerts upon the man's

hand a force equal and opposite to that which it exerts

upon the particle ;these two forces form the action and

reaction of which Newton speaks. It appears to the manthat the mass is trying to get away from his hand. For

this reason a force equal and opposite to the force neces-

sary to give the particle its normal acceleration has been

called "its centrifugal force." This is however a very

misleading term;

it seems to imply that the force belongs

to the mass instead of being an external force acting on

the mass. A somewhat less misleading term is "centri-

petal force." We shall avoid the use of either term;the

student who meets with them in the course of his readingwill understand that either means " the force which must

act on the mass to give it the acceleration normal to

the curve in which it moves."

EXAMPLES.

1. A particle, of mass m, moves on a horizontal table and is connected

by a string, of length 1, with a fixed point on the table ; if the greatest

weight that the string can support be that of a mass of M pounds, find the

greatest number of revolutions per second that the particle can make with-

out breaking the string.

Let n be the required number of revolutions so that the velocity of the

mass is n . '2x1.

.-. tension of the string=m .-

poundals.t

Hence Hg= mirhiH so that n = ?r (Mglml)*.Air

If the number of revolutions were greater than this number, the

tension of the string would be greater than the string could exert, and it

would break.

EXAMPLES. 249

2. 'A mass held by a string of length I describes a horizontal circle ;

if the tension of the string be three times the weight of the mass shew

that the time of a revolution is 2ir . /_ seconds.

3. With what number of turns per minute must a mass of 10

grammes revolve horizontally at the end of a string, half a metre in

length, to cause the same tension in the string as would be caused by a

mass of one gramme hanging vertically?

Ans. About 13-4.

4. A body, of mass m, moves on a horizontal table being attached to

a fixed point on the table by an extensible string whose modulus of

elasticity is X ; given the original length a of the string find the velocity

of the particle when it is describing a circle of radius r.

Ans.VI.

5. A particle whose mass is one-quarter of an ounce rests on a hori-

zontal disc and is attached by two strings, each four feet long, to the

extremities of a diameter. If the disc be made to rotate 100 times per

minute about a vertical axis through its centre, find the tension of each

string.

Am,. f|?r2poundals.

6. Two masses, m and m', are placed on a smooth table and con-

nected by a light string passing through a small ring fixed to the table.

If they be projected with velocities v and v' respectively at right angles to

the portions of the string, which is initially tight, find the ratio in which

the string must be divided at the ring so that both particles may describe

circles about the ring as centre.

Ans. In the ratio mi;2 : mV2.

7. A particle is moving in a circle of radius 200 feet with a speed, at

a given instant, of 10 feet per second. If the speed be increasing in

each second at the rate of half a foot per second find the resultant

acceleration.

Ans. . ft. -sec. unit of acceleration at an angle of 45 with thev^

tangent to the path.

250 THE CONICAL PENDULUM.

8. A skater, whose mass is 12 stone, cuts on the outside edge a circle

of 10 feet radius with uniformly decreasing speed just coming to rest

after completing the circle in five seconds ; find the direction and magni-

tude of his horizontal pressure on the ice when he is half way round.

1344Ans. =

o

tion of motion.

l\ + 47r2 poundals at an angle tan"1(2v) with his direc-

9. A wet open umbrella is held with its handle upright and made to

rotate about it at the rate of 14 revolutions in 33 seconds. If the rim of

the umbrella be a circle of one yard in diameter and its height above the

ground be four feet, shew that the drops shaken off from the rim meet

the ground in a circle of about five feet in diameter. If the mass of a drop

be -01 of an ounce shew that the force necessary to keep it attached to

the umbrella is about -021 of a poundal and is inclined at an angle tan"1

to the vertical.

161. The Conical Pendulum. If a particle be tied

by a string to a fixed point and move so that it describes

a circle in a horizontal plane, the string describing a cone

whose axis is the vertical line through 0, then the string

and particle together are called a conical pendulum.

THE CONICAL PENDULUM. 251

When the motion is uniform the relations between the

velocity of the particle and the length and inclination of

the string are easily found.

Let P be the particle tied by a string OP, of length I,

to a fixed point 0. Draw PN perpendicular to the

vertical through 0. Then P describes a horizontal circle

with N as centre [dotted in the figure].

Let T be the tension of the string, a its inclination

to the vertical, and v the velocity of the particle.

By Art. 156, the acceleration of P in the direc-

v*tion PN is -- and hence the force in that direction is

mI sin a

'

Now the only forces acting on the particle are the

tension, T, of the string and the weight, mg, of the

particle.

Since the particle has no acceleration in a vertical

direction we have

Tcosa = mg ..................... (1).

Also T sin a is the only force in the direction PN andrt

ffflfu

hence Tsina= 7 ; ......... ...(2).I sin a

From (1), (2) we have 7-^7-=-^-Zsm'a cos a

If the particle make n revolutions per second then

v = n . %7rPN = %7rnl sin a.

,or cos a = -p-^-r, ..... (3).^ 'cos a

252 MOTION OF A CARRIAGE ON A CURVE.

Also, by (1 ), T= 4w7rVZ poundals (4) .

.'. tension of the string : weight of the particle

The equations (3), (4) give a and T.

The time of revolution of the particle

_ 2-7T/ sin a/Tco&a.V-y *

and therefore varies as the square root of the depth of the

particle below the fixed point.

Hence it follows that for a governor of a steam engine

rotating 60 times per minute the height is about 978inches; for one making 100 revolutions per minute the

height is 3'58 inches; this latter height is too small for

practical purposes except for extremely small engines.

[For many interesting details on the subject of

governors of engines the student may consult Prof.

Kennedy's Mechanism, or books on the Steam Engine.]

162. Motion of bicycle rider on a circular path. When a man is

riding a bicycle on a curved path he always inclines his body inwardstowards the centre of his path. By this means the reaction of the

ground becomes inclined to the vertical. The vertical component of

this reaction balances his weight, and the horizontal component tendstowards the centre of the path described by the centre of inertia of the

man and his machine and supplies the necessary normal acceleration.

163. Motion of a railway carriage on a curved portion of the railwayline. When the rails are level the force to give the carriage the necessaryacceleration toward the centre of curvature of its path is given by the

action of the rails on the flanges of the wheels with which the rails are

in contact. In order, however, to avoid the large amount of friction that

would be brought into play and the consequent wearing away of the rails

the outer rail is generally raised so that the floor of the train is not hori-

zontal. The necessary inclination of the floor in order that there may beno action on the flanges may be easily found as follows.

MOTION OF A CARRIAGE ON A CURVE. 253

Let v be the velocity of the train, and r the radius of the circle

described by its centre of inertia G.

Let the figure represent a section of the carriage in the vertical plane

through the line joining its centre of inertia to the centre, 0, of the circle

which it is describing and let the section meet the rails in the points

A, B.

[The wheels are omitted for the convenience of the figure.]

Let R, S be the reactions of the rails perpendicular to the floor AB,and let be the inclination of the floor to the horizon.

The resolved part, (R + S) sin 6, of the reactions in the direction GO

supplies the force necessary to cause the acceleration towards the centre

of the curve.

.-. (R + S) sin e=m~. .(1).

Also the vertical components of the reactions balance the weight,

.-. (R + S) cos0 = m<7 (2).

v2From (1) and (2), tan =

, giving the inclination of the floor.

If the width AB be given we can now easily determine the height of

the outer rail above the inner.

It will be noted that the height through which the outer rail must be

raised in order that there may be no pressure on the flanges depends on

the velocity of the train. In practice the height is adjusted so that there

is no pressure for trains moving with moderate velocities. For trains

moving with higher velocities the pressure of the rails on the flanges

supplies the additional force required.

254 ROTATING SPHERE.

164. Rotating sphere. A smooth hollow sphere is rotating with

uniform angular velocity u about a vertical diameter; to shew that a

heavy particle placed inside will only remain resting against the side ofthe sphere at one particular level, and that if the angular velocity fallshort of a certain limit the particle will remain at the lowest point ofthe sphere.

Let AB be the axis of rotation of the sphere, A being the highest

point, and let be the centre;

let P be the position of the particle whenin relative equilibrium and PN the perpendicular on AB.

Now P describes a circle about N as centre with angular velocity u,

and therefore the force towards N must be mw2. PN or mu2a sin 6, where

a is the radius of the sphere and the angle POB.

The horizontal component of the normal reaction, R, at P supplies

this horizontal force and the vertical component balances the weight of

the particle.

Hence JJsin tf=mw2asin 6 ............................. (1),

Rco*6= mg ...................................... (2).

From equation (1) we have, either sin = 0, or R =

Substituting for I! in (2) we have

Hence the particle is either at the lowest point or at a point deter-

mined by equation (3).

The value of 6 given by (3) is impossible unless g < w2a, i.e. unless the

/ \*(-) Ifangular velocity is greater than -) If the angular velocity be less

than this quantity the only position of relative rest of the particle is at

the lowest point of the sphere.

165. Revolving string. A uniform inextensible endless string revolves

in the form of a circle, of radius r, on a horizontal smooth plane with

uniform angular velocity w about an axis through its centre perpendicularto its plane. If m be its mass per unit of length, to shew that its tension

Consider a portion, PQ, of the string which subtends a small angle 28

at the centre, 0. Let the tangents at P, Q meet in R so that QOR, ROPare each 6.

REVOLVING STRING. 255

The action of the rest of the string on the element PQ consists of the

two tensions, each T, along RP and RQ.

Their resultant is IT sin in the direction RO.

Now the smaller the arc PQ the more nearly is its motion the same as

that of a particle.

The mass of PQ is m . 2r0.

.'. the force towards must be 2mr0 . rw2.

Hence 2T sin 6= 2mr2w20, or T -~- = mr2w2

.

6

Now let 6 become indefinitely small ; then in the limit is unity,a

and we haveT=mr2w2

.

The above investigation includes the case of a revolving fly-wheel, and

gives also the minimum tension of a moving band which runs over a

revolving shaft.

EXAMPLES.

1. A string, of length four feet, and having one end attached to a

fixed point and the other to a mass of 40 pounds revolves, as a conical

pendulum, 30 times per minute ; shew that the tension of the string is

160ir2 ooundals and that its inclination to the vertical is cos"1! :

2. A heavy particle which is suspended from a fixed point by a string,

one yard long, is raised until the string, which is kept tight, makes an

angle of 60 with the vertical and is then projected horizontally in the

direction perpendicular to the vertical plane through the string ; find the

velocity of projection so that the particle may move in a horizontal

plane.

Am. 12 feet per second.

3. A particle, of mass m, is fastened by a string, of length I, to a

point at a distance b above a smooth table;

if the particle be made to

revolve on the table n times per second find the pressure on the table.

What is the greatest value of n so that the particle may remain in con-

tact with the table?

Ans. m(g- iirzn2b) poundals ; ^-

A/j-

256 EXAMPLES.

4. A particle is attached to a point A by an elastic string, whose

modulus of elasticity is twice the weight of the particle and whose natural

length is I, and whirled so that the string describes the surface of a cone

whose axis is the vertical line through A . If the distance below A of the

circular path during steady motion be I, shew that the velocity of the

particle must be*,/3</J.

5. Two masses, m and m', are connected by a string, of length c,

which passes through a small ring ; find how many revolutions per

second the smaller mass, ;//, must make, as a conical pendulum, in order

that the greater mass may rest at a distance a from the ring.

1 / in gAn*. - \/ } .

2ir V m'c-a

6. A railway carriage, of mass 2 tons, is moving at the rate of

60 miles per hour on a curve of 770 feet radius ;if the outer rail be not

raised above the inner, shew that the lateral pressure on the rails is equal

to the weight of about 1408 pounds.

7. A train is travelling at the rate of 40 miles per hour on a curve

the radius of which is a quarter of a mile. If the distance between the

rails be five feet find how much the outer rail must be raised above the

inner so that there may be no lateral pressure on the rails.

Ant. About !'. inches.

8. A mass is hung from the roof of a railway carriage by means of a

string, six feet long ; shew that, when the train is moving on a curve of

radius 100 yards at the rate of 30 miles per hour, the mass will movefrom the vertical through a distance of 1 foot 2 inches approximately.

9. A smooth right cone, whose vertical angle is 2a, with axis vertical

and vertex downward is made to revolve about its axis with uniform

angular velocity u;find where a particle must be placed on the inner

surface of the cone so that it may be in relative equilibrium.

An*. Its distance from the axis of the cone must be g cot a/or.

10. A smooth parabolic tube, whose latus rectum is 4a and vertex

downward, revolves uniformly about its axis, which is vertical. Shew

that, if the angular velocity be \/ ~ , a particle will rest anywhere,V iQj

whilst if the angular velocity differ from this quantity it will rest in no

position except the lowest point of the tube.

MOTION ON A CURVE. 257

11. A uniform circular string, of radius 10 feet and mass 1 lb.,

rotates uniformly about its centre 10 times per second ; shew that the

string will break unless its breaking tension is greater than the weight of

196 Ibs.

12. A thin circular hoop, six feet in diameter, is revolving in its

own plane about its centre. The ultimate tensible strength of iron is

60000 Ibs. per square inch of section and its density is 480 Ibs. per

cubic foot. Find the greatest number of revolutions per second that the

hoop can make without falling to pieces.

Ans. 40V10/7T.

166. The general case of the motion of a particle

constrained to move on a given curve under any givenforces is beyond the scope of the present book

;so also

is the motion of a particle constrained to move under

gravity on a given curve.

There is one proposition, however, relating to the

motion of a particle under gravity which we can prove in

an elementary manner and which is extremely useful for

determining many of the circumstances of the motion.

167. Theorem. If a particle slide down an arc of

any smooth curve in a vertical plane and if u be the

initial velocity and v its velocity after sliding through a

vertical distance h, to shew that v2 = u2 + 2gh.

M

L. D. 17

258 MOTION ON A CURVE

Let A be the point of the curve from which the

particle starts and B the point whose distance from A,

measured vertically, is h. Draw AM, BN horizontal to

meet any vertical line in M and N.

Let P, Q be two points on the curve very close to

one another and draw PR, QS perpendicular to MN.Then PQ is very approximately a small portion of a

straight line. Draw QV vertical to meet PR in V.

The acceleration at P along PQ is g cos VQP and

hence, if vp ,VQ be the velocities at P and Q, we have

vj =V + 2g cos VQP . PQ = vp* + 2g . VQ.

.'. VQ Vp = 2g . VQ, i.e. the change

in the square of the velocity is due to the vertical height

between P and Q. Since this is true for every element of

arc it is true for the whole arc AB. Hence the changein the square of the velocity in passing from A to B is

that due to the vertical height h so that v* = u* + 2gh.

168. The theorem in the preceding article may be

deduced directly from the Principle of the Conservation

of Energy.

For, since the curve is smooth, the reaction of the arc

is always perpendicular to the direction of motion of the

particle. Hence, by Art. 89, no work is done on the

body by the pressure of the curve. The only force that

does work is the weight of the particle.

Hence, since the change of energy is equal to the

work done, we have

^mu* = work done by the weight = mgh.

.-. v* = u> + 2h.

IN A VERTICAL PLANE. 259

169. If instead of sliding down the smooth curve the

particle be started along it with velocity u so as to move

upwards, the velocity v when its vertical distance from the

starting point is h is, similarly, given by the equation

= u -

Hence the velocity of the particle will not vanish until

it arrives at a point of the curve whose vertical heightu2

above the point of projection is ^- .

.

It will be noticed that the height to which the particle

will ascend is independent of the shape of the constraining

curve, nor need it continually ascend. The particle mayfirst ascend, then descend, then ascend again, and so on

;

the point at which it comes to rest finally will be at a

u*

height above the point at which its velocity is u.

t/

It follows that if a particle slide from rest upon a smooth

arc it will come to rest when it is at the same vertical

height as the starting point. An approximate example is

the Switch-back railway in which the car almost rises to

the same height as that of the point at which it started.

The slight difference between theory and experiment is

caused by the resistance of the air and the friction of the

rails which, although small, are not quite negligible.

The expression for the velocity when the particle is at

a vertical distance h from the starting point is the same

whether the particle be at that instant ascending or de-

scending.

170. The theorem is true not only of gravity but in

any case of the motion of a particle on a smooth curve

under the action of a constant force in a constant direc-

172

260 NEWTON'S EXPERIMENTAL LAW.

tion, e.g. in the case of motion on a smooth inclined plane,

if we substitute for"g" the acceleration caused by the

forces. It is also true if we substitute for the constraining

curve an inextensible string fastened to a fixed point, or

a weightless rod which is always normal to the path of the

particle.

We cannot, in general, find the time of describing any

given arc without the use of the Differential Calculus.

171. Newton's Experimental Law. By using the

theorem of Art. 167 we can shew how Newton arrived at

his law of impact as enunciated in Art. 141.

We suspend two spheres, of small dimensions, by

parallel strings whose lengths are so adjusted that when

hanging freely the spheres are just in contact with their

centres in a horizontal line.

One ball, A, is then drawn back, the string being

kept tight, until its centre is at a height h above its

original position and then allowed to fall. Its velocity

v on hitting the second ball B is Jfyh.

Let v, v" be the velocities of the spheres immediately

after the impact, and h', h" the heights to which they rise

before again coming to rest so that

,and v" =

The sphere A may either rebound, remain at rest, or

follow after B.

Taking the former case, the relative velocity after

impact is -(v' + v") or - Jig (*Jh' + *Jh").

Also the relative velocity before impact was J2g . \/h.

We should find that the ratio of (*Jhf

-f \/h") to *Jh

would be the same whatever be the value of h and the

ratio of the mass of A to that of B, and that it would

MOTION IN A VERTICAL CIRCLE. 261

depend simply on the substances of which the masses

consist.

We have only considered one of the simpler cases. Bycarefully arranging the starting points and the instants of

starting from rest both spheres might be drawn aside and

allowed to impinge so that at the instant of impact both

were at the lowest points of their path. The law enun-

ciated by Newton would be found to be true in all cases.

172. Motion on the outside of a vertical circle. A particle slides

from rest at the highest point down the outside of the arc of a smooth

vertical circle ; to shew that it will leave the curve when it has described

vertically a distance equal to one third of the radius.

Let be the centre and A the highest point of the circle. Let v be

the velocity of the particle when at a point P of the curve, R the pressure

of the curve there, and r the radius of the circle. Draw PN perpendicu-

lar to the vertical radius OA and let AN=h.

Then v2= 2g . AN=2gh.The force along PO is mg cos 6 -R.

v2But the force along PO must, by Art. 156, be m . .

r

vz

. . m mg cos - R,r

2~] r r-h 2gh~\--J=m ^

----^-J

r-3h= mg .

7*

Now R vanishes and changes its sign when 37z= r, or when h= -.o

The particle will then leave the curve and describe a parabola freely ; for

to make it continue on the circle the pressure R would have to become

a tension; but this is impossible since the curve cannot pull the

particle.

Cor. If the curve, instead of being a circle, be any other curve in a

vertical plane the particle will leave it similarly at a depth h where

p being the radius of curvature at that depth and 6 its inclination to

the vertical.

262 MOTION IN A VERTICAL CIRCLE.

173. Motion in a vertical circle. A particle, of mass m, is susp<'>i<l<'<l

by a string, of length r, from a fixed point and hangs vertically. It is then

projected with velocity u so that it describes a vertical circle ; to find the

tension and velocity at any point of the subsequent motion and to find also

the conditions that it may (a) make complete revolutions, (j3) oscillate, and

(y) cease moving in a circle.

mff

Let be the point to which the string is attached, OA the vertical

through A and OC the horizontal radius.

Let v be the velocity at any point P of its path and T the tension of

the string there.

Let PN be drawn perpendicular to OA, let AN=h, and let the angle

POA be 6.

Then, by Art. 167, v*=u*-2gh (1).

Also, by Art. 156,

m - = force at P along the normal PO to the path of the particle.r

v* r-h.-.TO = T- mg cos ff=T-mg ,

r r

(2).

MOTION IN A VERTICAL CIRCLE. 263

These two equations give the velocity and tension at any point.

(o) The particle will make complete revolutions if the velocity do

not vanish before it reaches the highest point B of the path and if, in

addition, the value of the tension given by (2) do not become negative.

This latter condition is necessary; for otherwise, in order that the

circular motion might continue, the pull of the string would have to

change into a push and this is impossible in the case of a string.

It follows from (2) that T diminishes as h increases;hence if it be

positive at the highest point B, where h=2r, it is positive everywhere.This will be the case if w2 + g (r

-6r) be positive, or if u2 be > 5gr.

Also, if w2 be > 5gr, the value of v given by (1) does not vanish any-where and the particle makes complete revolutions.

If w2 be < 5gr the tension vanishes before the particle reaches the

highest point and two cases arise.

The particle may rise to a certain height and then retrace its path, or

it may leave the circular path altogether.

The first of these two cases arises when the velocity v vanishes whilst

the tension still remains positive ;the latter case when the tension

vanishes before the velocity.

(/3) From (1) the velocity vanishes at a height w2/2# above the

lowest point, and from (2) the tension vanishes at a height (u?+ gr)[3g.

Hence the velocity vanishes before the tension if

M2/2# be < (u2 + gr)j3g, i.e. if u2 be < 2gr,

i.e. if the velocity of projection be insufficient to take the particle as highas the centre 0.

Hence the particle will oscillate if w2 be < 2gr.

If it2= 2gr the tension will vanish at the same time as the velocity

but, as it has not become negative, the motion will still be oscillatory.

(y) If w2 be > 2gr, then u-j2g is > (u? + gr)j3g, and the tension will

vanish before the velocity at a height (w'J + gr)/3g, which is greater than r.

Hence the circular motion will cease at some point of the path between

C and B;the string will become slack and the particle will describe a

parabola freely.

To sum up the results of this investigation.

If w be < i>j2yr the particle oscillates in an arc less than a semi-

circle.

If u= ijlgr the arc of oscillation is a semi-circle.

If u be > rjfyr and < *Jogr the particle ceases describing the circle

somewhere between G and B.

264 MOTION IN A VERTICAL CIRCLE.

If = j5gr the particle just makes complete revolutions.

The results of this article apply without any alteration if the particle

move on the inside of a metal curve in the shape of the above circle. Wehave only to substitute the pressure of the curve for the tension of the

string.

Cor. The maximum value of the tension is at the lowest point and

its value there is m (uz+ gr)jr; also if the particle makes complete oscilla-

tions u t j5gh.

.. the tension is not < 6mg, and hence the breaking tension of the

string must be at least six times the weight of the particle.

174. If the motion be inside a metal tube in the shape of a circle the

investigation of the motion is similar to that of the last article ; in this

case, however, the pressure can change its sign ; for the particle may pass

from contact with the one surface of the tube to contact with the other.

Hence the particle will make complete revolutions if the velocity of pro-

jection be sufficient to carry it to the highest point, i.e. if u be <t ijigr ;

otherwise it will oscillate.

Ex. 1. A particle slides down the arc of a vertical circle; shew

that its velocity at the lowest point varies as the chord of the arc of

descent.

Ex. 2. A heavy particle is attached by a string, 10 feet long, to a

fixed point and swung round in a vertical circle. Find the tension and

velocity at the lowest point of the circle so that the particle may just

make complete revolutions.

Am. Six times the weight of the particle ;40 feet per second.

Ex. 3. If a particle move in a smooth vertical circular tube, of radius

a, with velocity due to a height h above the lowest point of the circle,

prove that the pressure at any point will be proportional to the depth

below a horizontal line at a height above the lowest point.o

Ex. 4. A particle is projected along the inside of a smooth vertical

circle, of radius a, from the lowest point ;find the velocity of projection

so that after leaving the circle the particle may pass through the

centre.

Ans.

APPARENT VALUE OF GRAVITY. 265

Ex. 5. A heavy particle is fastened to the middle point, C, of a

string ACB, two yards in length, the ends of which, A and B, are

fastened to fixed points in the same horizontal line at a distance 3^/3feet from each other. The particle, when hanging at rest, has given to

it a velocity of 16 feet per second perpendicular to the plane ACB ; shew

that it will completely describe a vertical circle and that the greatest and

least tensions of the string will be respectively nineteen-thirds and one-

third of the weight of the particle.

175. Effect of the rotation of the earth on the apparent

weight of a body. The earth rotates once per day on its

axis so that any particle P on its surface describes a circle

round the foot N of the perpendicular drawn from P to

the axis. Hence there must be some force in the direction

PN. Now the forces acting on a particle resting on the

earth's surface are (1) the attraction of the earth on the

particle, and (2) the pressure of the ground on it. The

resultant of these two forces must be in the direction of

the line PN. This pressure of the earth on the body is

the apparent value of gravity and is the same as the

tension of the string which would support the body if

hanging freely.

176. Assuming the earth to be a sphere of radius a

feet, to find the apparent value and direction of the attrac-

tion of the earth at any point of its surface taking into con-

sideration the revolution of the earth about its axis.

Let P be a point in latitude X, PN the perpendicularon the axis, OA, of the earth, and APB the meridian

through P, so that the angle POB is X. Also let the

angular velocity of the earth be &>.

Let ra be the mass of a particle at P, mg the attraction

of the earth on it directed towards the centre 0, and let

Y, X be the components of the pressure of the earth on

266 EFFECT OF THE REVOLUTION OF THE EARTH

the particle along and perpendicular to the radius OPrespectively.

Since P describes, with uniform angular velocity to, a

circle whose centre is JV, its only acceleration is w'2 .PN

directed along PN.Hence resolving the forces X, Y, and mg along and

perpendicular to PN we have

X sin X + (mg Y) cos X = raw2. PN= mafa cos X . . .(1),

Xco*\-(mg- F)sinX = (2).

Solving these equations we have

X = ma>*a cos X sin X, and Y= m (g w2 a cos2

X).

Assuming as a rough approximation that the earth

turns on its axis once in 24 hours and that the radius of

the earth is 4000 miles, we have

6)*o. _ / 2?r

~g~~

V24 x 60 x 60

2 4000 x 1760 x 3 , 1x a* about

289'

UPON THE APPARENT VALUE OF GRAVITY. 267

Hence is a small quantity and so its square maybe neglected.

Let the resultant of X and Y be G at an angle 6 with

the radius OP.

Then G = JX*+Y*

= mg\(\ cos2

Xj +( cosXsinXJ >

(?\

1 cos2 X

j,

at a(by the Binomial Theorem, neglecting the square of

. y= mg ma?a cos

2X.

X a?a cos X sin X &>2a .

Also tan = ^ =5

= sin X cos X.J # &) a cos X #

Hence the effect of the rotation of the earth is to

diminish the apparent weight by ma?a cos2

X, and to turn

its direction through an angle

.t/Va . .\

tan - sin X cos X .

At the equator the apparent weight is less than the

real weight by mafa; if we take accurate values of the

quantities involved this diminution is found to be about

of the real weight.

Cor. The apparent pressure at the earth's surface at

the equator would just vanish (i.e. the attraction of the

earth would be just sufficient to keep a particle rotatingwith the earth) if Y were zero there, i.e. if the angular

velocity to of the earth were A/-. If the angular velocityV CL

268 EXAMPLES ON CHAPTER VII.

of the earth were to exceed this quantity particles would

not remain in contact with the earth unless compelled to

do so.

This limiting angular velocity is roughly

32r

4000x1760x3 100,/66'

The time of the earth's revolution would then be

1~100 V66

seconds or about 1 hour 25 minutes.

Hence if the earth were to rotate about 17 times faster

than at present bodies at the equator would leave its

surface.

Ex. If the earth were set to revolve on its axis in half a day,

nothing else being altered, find what would be the apparent weight of a

mass of 100 pounds, as tested by a spring balance, supposing the

balance to be graduated to shew the weight of pounds at the equa-

tor, and assuming that the earth's radius is 21 x 106 feet and that

the value of g is 32-09 at the equator with the actual angular

velocity.

Ans. Wt. of 98-96 Ibs.

EXAMPLES. CHAPTER VII.

1. A railway carriage is travelling on a curve of radius

r with velocity v; if A be the height of the centre of inertia of

the carriage above the rails and 2a the distance between them,

shew that the weight of the train is divided between the inner

and outer rails in the ratio of yra v2h to gra + v*h, and hence

that the carriage will upset if v be >AT-

EXAMPLES ON CHAPTER VII. 2G9

2. A particle, of mass m, is at the centre of a fine elastic

string, of natural length 2, the ends of which are held in a

man's hands at a distance 2a apart. If the mass be then

swung round with angular velocity CD, shew that the angle

which the portions of the string make with the line joining

the hands is 2sin~M-^*/ ],where A is the modulus of

elasticity of the string.

3. A bead is strung on a thread whose two extremities

are fixed;shew that if it be projected so as to move on a

horizontal plane passing through the two points then the

tension of the string varies inversely as the product of the

two portions of the string joining it to the fixed points.

4. A leather belt, whose mass is 3 Ibs. per foot of its

length, is running over a pulley with a velocity of 100 feet

per second. Find the least tension permissible in the belt

and shew that it is independent of the radius of the pulley.

5. A circular elastic band is placed about a wheel whose

circumference is twice the natural length of the string and the

wheel revolves with constant angular velocity ;find the pres-

sure of the band on the wheel.

6. Shew that at the equator a shot fired westward with a

velocity of 8333 metres per second, or eastward with 7407

metres per second, would, if unresisted, move horizontally

round the earth in one hour and twenty minutes and one hour

and a half respectively.

7. A man stands at the North Pole and whirls a mass

of 24 Ibs. Troy on a smooth horizontal plane by a string, one

yard long, at the rate of 100 turns per minute; he finds that

the difference of the forces which he has to exert according as

he whirls one way or the other is roughly equal to the weightof 39 grains ;

find the period of rotation of the earth.


8. A railway train, of mass m, is travelling on a smooth

line with velocity v along a parallel of latitude in latitude \;

shew that the difference in the normal pressure according as

it is going east or west is 4mvw cos A poundals, where o> is

the angular velocity of the earth.

9. If the earth were to rotate so fast that particles on its

surface at the equator were just on the point of leaving it,

prove that the greatest inclination of the plumb-line at any

point to the radius of the earth through the point would be

tan" 1

J.

10. A string, of length I, has its ends fastened to two

points, A and Ji, in the same vertical line and a bead P on the

string rotates about AB with a uniform angular velocity o>.

If w be less than {2gl/(P-a*)fi shew that it will hang

vertically, and that BP will be horizontal if <o be equal to

l{2g/a(P -a')}*.

11. A string, of length a, is fastened to a point A andcarries a mass P. If P be projected vertically upwards whenAP is below the horizontal line and makes an angle a witli it,

find the least velocity of projection that P may ultimatelydescribe a circle.

12. A particle hangs vertically by a string and is

projected horizontally and rises to P;

it then leaves circular

motion and commences circular motion again at Q ; prove that

PQ and the tangent at P make equal angles with the vertical.

13. A string passing through a small hole in a smooth

horizontal table has a small sphere, of mass m, attached to

each end of it;the upper sphere revolves in a circle on the

table when suddenly it strikes an obstacle and loses half its

velocity ;find what diminution must be made in the mass of

the lower sphere so that the upper one may continue rotating

in a circle.

EXAMPLES ON CHAPTER VII. 271

14. A string PAQ passes through a hole A in a smooth

table the portion AP lying on the table and AQ being at an

angle of 45 to the vertical and below the table, so that P and

Q are in the same vertical line. If masses be attached at Pand Q and, the strings being stretched, be each projected

horizontally, find the velocity of projection and the ratio of the

masses so that the plane PAQ may always be vertical and the

angle PAQ always 45. If the string be four feet in length

find the time of revolution.

15. Two particles, of masses m and m, are connected byan inextensible string of length a and lie on a smooth table at

a distance a from each other. The particle of mass m is

projected at right angles to the string with velocity v. Shew/ g

that the tension of the string is always . .m + m' a

16. A particle is projected horizontally from the highest

point of a parabola whose axis is vertical and vertex upwards ;

if the particle ever leave the curve at all it will do so at

the instant of projection.

17. A smooth parabolic curve is fixed with its plane

vertical and axis horizontal;

shew that a heavy particle

placed on the arc at a height above the axis equal to the latus

rectum will run off the curve on arriving at the end of the

latus rectum and then describe a parabola having the same

latus rectum as the given arc.

18. A bead slides on a wire bent into the form of a

parabola whose axis is vertical and vertex upwards ;if the

bead be just displaced from its position of equilibrium, shew

that in any subsequent position the pressure on the curve

varies as the curvature of the curve, and the velocity of

the bead varies as its distance from the axis of the parabola.


1 9. Particles start from all points of the arc of a parabola,whose latus rectum is 4a and whose axis is vertical, and slide

down to the vertex arriving there simultaneously. They then

proceed to fall freely ;shew that after a time . / each is atV g

the same distance from the axis that it was initially.

20. An ellipse is placed with its minor axis vertical and a

smooth particle slides down it starting from rest at the highest

point and leaves the curve at the point whose eccentric angleis

<fr.Shew that the eccentricity of the ellipse is

72-3sii

V 2 + sin1 - sin<f>V 2 + sin <

21. An elliptic wire ofeccentricity./^

is placed in a

vertical plane with its major axis inclined to the horizon at an

angle of 60. If a small ring be allowed to slide along the wire

from the higher extremity of the major axis shew that its

velocities at the two ends of the minor axis are as J2 : 1.

22. An ellipse in a vertical plane has its major axis

horizontal and a particle is constrained to describe the circle

of curvature at its lowest point starting from rest at the

highest point of the circle; at the lowest point it is gently

guided inside the ellipse; shew that the particle will com-

pletely describe the ellipse if its eccentricity be greater

than ^.

23. Two equal particles at rest are connected by a fine

inextensible string of length 3a. One particle is free to moveon a horizontal table in a smooth parabolic groove whose latus

rectum is 4a. The string passes through a hole in the table

at the centre of curvature of the vertex of the groove and

initially the second particle is just at the level of the table.


It then falls freely under gravity ;shew that the first particle,

when it is passing through the vertex of the groove, exerts no

pressure on the sides of the groove.

24. Shew that a particle of mass m may describe a

parabola, whose focus is S and vertex A, with uniform

velocity V under the action of two equal forces, one attractive

towards S and the other repulsive and perpendicular to the

directrix, provided that the forces are each inversely propor-

tional to the distance of the particle from the focus and

72are each equal to m -

j when the particle is at the vertex.

25. A particle is describing uniformly a horizontal circle

within a hollow bowl in the shape of a prolate spheroid whose

axis is vertical;shew that the time of a complete revolution

varies as the square root of the depth of the particle below the

centre of the spheroid.

26. A particle slides from rest at the vertex down the arc

of a smooth vertical circle ; find the equation of its hodograph.

27. A ball of mass m is just disturbed from the top of a

smooth circular tube in a vertical plane and falls impinging on

a ball of mass 2m at the bottom, the coefficient of restitution

being i;find the height to which each ball will rise in the

tube after a second impact.

28. A particle at the end of a fine string just makes

complete revolutions in a vertical circle. If the particle were

five times as heavy as it is the string would just bear the

greatest strain. At the highest point of its path another

particle of five times its mass and moving in the same circle

with five times its velocity overtakes it; given that the string

is now on the point of breaking when the tension is greatest

shew that the modulus of elasticity of the particles is equal

to*.

L. D. 18


29. A light inextensible string has one extremity fixed, a

heavy particle attached at the other extremity, and another

heavy particle attached at an intermediate point ;the particles

describe horizontal circles with the same angular velocity o>.

Shew that the inclinations of the two parts of the string to the

vertical are cot" 1

^/wVj and cot~ l

gl<i>'r3 ,where r^ and r.z are the

radii of the circles described by the centre of inertia of the

two particles and the lower particle respectively.

30. Watt's Governor consists of a rhombus of jointed rods

attached at one angle to the top of a vertical revolving shaft

and at the opposite angle to a collar free to slide up and down

the shaft and carrying two heavy masses at the angles, each

equal to M. Supposing the masses of the rods to be neglected

in comparison with M, find the angle to which the rods open

as the shaft revolves and find the change produced in this

angle by attaching a load M' to the collar.

31 . Two particles, of masses m and in, are fixed to the ends

of a weightless rod, of length 21, which is freely moveable about

its middle point. Shew that the inclination a of the rod to

the vertical when the particles are moving with uniform

angular velocity w about a vertical axis through the centre of

the rod is given by the equation (m + m) <aal cos a = (m m') g.

32. Assuming that the planets move in circles about the

sun as centre under an attraction to the sun which varies

inversely as the square of the distance, shew that the squares

of their times of revolution vary as the cubes of their dis-

tances.

33. A gun is suspended freely by two equal parallel cords

so that its inclination to the horizon is a and a shot whose

mass is -th that of the gun is fired from it. If h be then


height through which the gun recoils shew that the rangeof the shot on the horizontal plane through the muzzle is

11 (1 + n) h tan a.

34. A loaded cannon is suspended from a fixed horizontal

axis and rests with its axis horizontal and perpendicular to the

fixed axis, the supporting ropes being equally inclined to the

vertical;

if v be the initial velocity of the ball whose mass is

-th that of the cannon and h be the distance between the axisn

of the cannon and the fixed axis of support, shew that, when

the cannon is fired off, the tension of each rope is immediatelyaltered in the ratio v2 + n2

gh : n (n + 1) gh.

If the cannon were supported in a gun-boat in the manner

described with its axis in the direction of the boat's lengthwhat would be the effect of firing it 1

35. Two equal particles are connected by a string, of

length Tra, and rest symmetrically over a smooth vertical circle

of radius a. If they be just displaced from the position of

equilibrium, shew that the upper particle will leave the arc

when the radius to it from the centre makes an angle with

the vertical given by (0 + -\-1 cos 6.

36. A lamina in the form of a regular hexagon, whose

side is a, is placed on a smooth table and fixed to the table. Astring, whose length is equal to the perimeter of the lamina, is

wound round it, one end being attached to an angular point,

and carries a mass ra attached to its other end. If the

particle be projected, with velocity u, horizontally at right

angles to the string, find the time that elapses before the

string is again wound up and determine the greatest and least

tensions of the string.

182


37. A prism (whose transverse section is a regular

hexagon) is held with its axis horizontal and a string equal

in length to the perimeter of the hexagon with a heavy

particle at one end is suspended by the other end from a point

in the lowest edge of the prism which is in the same vertical

plane as the axis. Find the least velocity with which the

particle must be projected at right angles to this plane so that

the string, which is always kept tight, may be drawn com-

pletely round the prism.

38. Two masses, m and m', are connected by an inexten-

sible string of length a. The extremity A, to which m is

attached, is compelled to move with uniform acceleration

f in a straight line under a force P in that line, and the

extremity ,to which m' is attached, is compelled to describe

a circle round A with uniform angular velocity o> under the

action of a force Q perpendicular to AB. Find P and Q, and

,

shew that the least value of P is mf--j> , provided that

aw2 is < 2f.

39. Two particles of equal mass are joined by a rod

without weight and are placed in a horizontal position on

a plane inclined to the horizon at an angle a; the plane is

smooth with the exception of a rough spot on which one of

the particles, A, rests\the coefficient of friction being p tan a,

shew that, if p be < 4, the particle A will begin to move wheii

the rod has turned through an angle sin" 1^-TTT-

What happens if p be > 4

CHAPTER VIII.

SIMPLE HARMONIC MOTION. CYCLOIDAL AND PENDULUMMOTIONS.

177. Theorem. If a point Q describe a circle with

uniform angular velocity and ifP be the foot of the perpen-

dicular drawn from Q upon a fixed diameter AOA' of the

circle, to shew that the acceleration of P is directed towards

the centre, O, of the circle and varies as the distance of P

from 0, and to find the velocity of P and its time of

describing any space.

278 SIMPLE HARMONIC MOTION.

The velocity and acceleration of P are the same as

the resolved parts, parallel to AO, of the velocity and

acceleration of Q.

Let G> be the constant angular velocity with which the

point Q describes the circle.

Let a be the radius of the circle and let the angle

QOA be 0. Draw QT the tangent at Q.

By Art. 156, Cor. 1, the acceleration of Q is ao>*

towards 0.

.: the acceleration of P along PO = aeoacos 8 = <u* . OP,

and therefore varies as the distance of P from the centre

of the circle.

Also the velocity of P= aa> cos QTO = ao> sin 6 = o> . PQ = (aja* of,

where OP is x.

This velocity is zero at A and A', and greatest at 0.

Also the acceleration vanishes and changes its sign as

the point P passes through 0.

The point P therefore moves from rest at A, has its

greatest velocity at 0, comes to rest again at A', and then

retraces its path to A.

Also the time in which P describes any distance AP= time in which Q describes the arc AQ

- = - COS\a

7T/. time from A to A' = -

.

eo

2-7T

Also time from A to A' and back again to A = .

178. Simple Harmonic Motion. Def. If a point

move in a straight line so that its acceleration is always

SIMPLE HARMONIC MOTION. 279

directed towards, and varies as its distance from, a fixed

point in the straight line the point is said to move with

simple harmonic motion.

The point P in the previous article moves with simpleharmonic motion.

From the results of the previous article we see, by

equating o>'2to

//,,that if a point move with simple harmonic

motion, starting from rest at a distance a from the fixed

centre 0, and moving with acceleration//. . OP then

(1) its velocity when at a distance # from is

(2) the time that has elapsed when the point is at a

distance x from is -7- cos" 1 -

VA6 a

(3) the time that elapses before it is again in its

j... , ... . 27T

initial position is r .

179. The range, OA or OA', of the moving point on

either side of centre is called the Amplitude of the

motion.

The time that elapses from any instant till the instant

in which the moving point is again moving through the

same position with the same velocity and direction is called

the Periodic Time of the motion.

2-7TIt will be noted that the periodic time, -.

,is inde-

vppendent of the amplitude of the motion.

180. The Phase of a simple harmonic motion is the fraction of the

whple period that has elapsed since the point last passed through its

extreme position in the positive direction.

280 SIMPLE HARMONIC MOTION.

a

Thus when the particle is at P the phase is ^- . T, where T is the

periodic time.

If the time be measured from the instant at which the corresponding

point Q is at E the angle EOA is called the Epoch and is generallydenoted by e.

If P be the position of the moving point at time t and T be the

periodic time the angle EOQ is 2-r = , and hence the distance OP

/2ir \I -~ t-e \ or a cos (/Jut

-e).

181. It BOB', Fig. Art. 177, be the diameter perpendicular to AOA'and QF be drawn perpendicular to it, the point P' will have a simpleharmonic motion whose amplitude and period are the same as those of Pbut whose phase is less than that of P by one-quarter of a period ; for

when Q is at 11 the phases are respectively zero and T.

Hence we see that a point which possesses two rectangular simple

harmonic motions of the same period and amplitude, but of phases

differing by one-quarter of a period, moves in a circle whose radius is

the common amplitude.

182. Two simple harmonic motions in the same straight line, of the

same period, give a harmonic motion of the same period. For the

distance of the point must =a cos (Jut-

e) + a' cos (Jut-

e')

where a" cos e" = a cos t + a' cos e', and a" sin e"= a sin e+ a' sin e'.

a sin e +a' sin e'

Hence a"= Ja~+ a'2+ 2aa' cos (e-

e') and tan e"=; , ,

a cos e + a cos e

so that the new amplitude is the diagonal of a parallelogram whose sides

are the original amplitudes inclined to one another at an angle equal to

the difference of the epochs.

183. The time of vibration in simple harmonic motion

may, by consideration of dimensions only, be shewn to be

independent of the extent of oscillation. For, since the

acceleration at any point is /* x a distance, the dimensions

of /* must be 2 in time. Now the time of a complete

SIMPLE HARMONIC MOTION. 281

oscillation can only depend on//,

and a. Let it equal

y? . av so that its dimensions are [7T

]~2a;

[Z]t

'. Hence as = \

and y = 0, so that the required time varies as -r- and is

V/ti

independent of a.

184. Extension to motion in a curve.

O' P' B

Suppose that a moving point P is describing a portion,

AOA', of a curve of any shape starting from rest at Aand moving so that its tangential acceleration is always

along the arc towards and equal to //. . arc OP, then the

preceding proposition is true with slight modifications.

For let O'B be a straight line equal in length to the

arc OA and let P' be a point describing it with accelera-

tion n . O'P';also let O'P' = arc OP.

Since the acceleration of P' in its path is always the

same as that of P, the velocities acquired in the same

time are the same and the times of describing the same

distances are the same.

Hence

(1) The velocity of P- O'P'

2

)=

fi {(arc OA )2 -

(arc OP)2

},

282 EXAMPLES.

(2) Time from A to P1 -l fO'P\ I ../arcOPN=

-7- COS n,-^-= COS TT-J- ,

\V \OBJ *Jfj, \dircOAJ

2?r(3) Time from A to -4' and back again = .

^A*

EXAMPLES.

1. A point, moving with simple harmonic motion, has a velocity of

4 feet per second when passing through the centre of its path and its

period is TT seconds; what is its velocity when it has described one foot

from the position in which its velocity is zero?

Aw. 2^3 feet per second.

2. A mass of one gramme vibrates through a millimetre on each side

of the middle point of its path 256 times per second ; assuming its

motion to be simple harmonic, shew that the maximum force upon the

particle is ^ (512ir)- dynes.

3. A horizontal shelf moves vertically with simple harmonic motion

whose complete period is one second;

find the greatest amplitude in

centimetres that it can have so that objects resting on the shelf mayalways remain in contact with it.

Ans. Nearly 25 centimetres.

4. A particle is oscillating in a straight line about a centre of force,

0, towards which when at distance r the force is n2r, and a is the ampli-

/gtude of the oscillation ; when at a distance a ~- from the particle re-

2

ceives a blow in the direction of motion which generates a velocity na.

If this velocity be away from 0, shew that the new amplitude is a^/3.

5. A particle is initially at a distance c from a fixed point and

its acceleration is directed towards, and is equal to /j. times its distance

from the fixed point ; shew that it will arrive at the fixed point in time

j- sin"1.

, where V is the velocity of projection.

THE CYCLOID. 283

''. An elastic string is extended between two fixed points to twice its

natural length and a particle, of mass m, is fastened to the middle pointof the string. The particle is drawn towards one fixed point through a

distance equal to half its distance from it and then let go. Find the

greatest velocity subsequently attained and, if a be the natural length of

the string, shew that the time of an oscillation is T A . where X is

the modulus of elasticity of the string.

7. One end of an elastic string, whose modulus of elasticity is X, is

fixed to a point on a smooth horizontal table and the other end is tied to

a particle, of mass m, which is lying on the table. The particle is pulledto a distance from the point of attachment of the string equal to twice

the natural length, a, of the string and then let go ; shew that the time of

a complete oscillation is 2(ir + 2) . /-.

am\

'

8. An elastic string without weight, of which the unstretched lengthis / and the modulus of elasticity the weight of n ozs., is suspended byone end and a mass of m ozs. is attached to the other; shew that the

time of a vertical oscillation of the mass is 2ir . / .

n g

9. Find the resultant of two simple harmonic vibrations in the samedirection and of equal periodic time, the amplitude of one being twice

that of the other and its phase a quarter of a period in advance.

Ana. A simple harmonic vibration of amplitude V5 times that of the

first and whose phase is in advance of the first by (tan-1

2)/2ir of a

period.

Motion on a Cycloid and Pendulums.

185. In the following articles we shall discuss the

motion of simple forms of Pendulums. Intimately con-

nected with this motion however is motion down a smooth

wire in the form of a curve called a Cycloid. It will be

desirable to define and call attention to its properties.

284 GEOMETRICAL PROPERTIES

186. Cycloid. Def. If a circle roll, without sliding,in one plane on a fixed straight line the path of any point

fixed on the circumference of the circle and moving with it

is called a Cycloid.

[It is the path of a speck of mud on the edge of a cartwheel which is

rolling in a straight line on a level road.]

Thus let the circle APB roll on the straight line

C'A'C so that a point P fixed on its circumference

describes the arc C'B'PG. This curve is a portion of a

cycloid.

Let A'P'B' be the position of the circle when the fixed

point is at its greatest distance from CC'.

Then B' is called the vertex, A'B' the axis, and C'C

the base of the cycloid.

If a perpendicular from P on A'B' meet the circle

A'P'B' in P' then

(1) The tangent to the cycloid at P is parallel to

B'P',

(2) The arc B'P = twice the line B'P', and

(3) The radius of curvature at P is equal to twice

the line AP, i.e., is equal to twice the portion of the

normal intercepted between the curve and the base.

<)F THE CYCLOID. 285

In the following articles we shall give proofs of these

propositions. For other proofs the student may refer to

Thomson and Tait's Natural Philosophy, Vol. I., Art. 93,

or to Edwards' Diffei'ential Calculus, Art. 356 etc.

187. Let Q be a consecutive point on the cycloid and let perpendiculars

from P and Q upon A'O'B' meet the circle A'P'B' in P and Q'. Also

draw l*N perpendicular to B'Q'.

We have shewn (Art. 40, Ex. 1) that the point P is for the instant

turning about the point . ! with the same angular velocity with which the

circle is turning about its centre.

Hence the point P is moving perpendicular to AP and the tangent

at P is therefore UP which is parallel to B'P'.

Again since P has the same angular velocity about . I that the circle

has about its centre and since, whilst the circle turns through a small

angle P'O'Q', P advances to Q,

Now P'B'Q' is very small and B'P'N, B'NP' are ultimately right

angles so that B'N=B'P' ultimately.

Hence the increase in the arc B'P is equal to twice the increase in the

line B'P';

also the arc B'P and the line B'P' vanish together at the

vertex ; hence the arc B'P is twice the line B'P'.

It follows that the whole arc B'C is twice B'A' or four times the

radius of the rolling circle.

188. Radius of curvature. Since B'P', B'Q' are parallel to the

tangents at P and Q, the angle between these tangents is P'B'Q' or

Hence the radius of curvature at P= Lt. arc PQ-r- angle between the tangents at P and Q= Lt. 2 arc FQ' cos 0-=- i / P'O'Q'

= 4 cos e x Lt. arc P'Q'+L P'O'Q'

286 MOTION ON A VERTICAL

189. Let C'B'C be a cycloid with axis vertical and vertex downward.

X,

Draw the generating circle APB in the position in which it intersects

the cycloid in P, so that AP is the normal at P.

Produce B'A' to E' making A'E' equal to A'B' and draw E'E parallel

to A'A to meet BA produced in E ; also produce PA to Q making AQequal to PA.

Then since the triangles BAP, EAQ are equal in all respects, EQA is

a right angle and a circle on EA as diameter will pass through Q, so that

arc EQ = arc BP= line A'A- E'E.

Hence if Q be a fixed point on the circle EQA, and if the circle,

starting with Q in contact with E'E at E', roll along E'E the fixed pointwill trace out a portion of a cycloid, E'QC, having its cusp at E' and

vertex at C.

Also the arc QC=2. line AQ = PQ.

Hence it follows that if a string be wrapped tightly round the curve

E'QC, one end being fixed at E' and the other being initially at C, this

end will when the string is unwrapped describe the arc CPB'; if the

string then wrap itself upon a similar portion of a cycloid, having vertex

at C" and cusp at E', its end will describe the other portion B'C' of the

original cycloid.

The student who is acquainted with the Differential Calculus will

observe that Q is the centre of curvature of the curve at P and that the

arc E'QC is therefore a portion of the evolute of the original curve, so

that the above proposition is a particular case of the proposition which

is true for all curves and their evolutes.

CYCLOID UNDER GRAVITY. 287

190. Theorem. If a smooth cycloid be placed in a

vertical plane, with its axis vertical and its vertex downward,to shew that the time of oscillation of a particle from rest

to rest is the same whatever be the point of the curve fromwhich the particle starts.

Let (JHC be the cycloid, CC' being the base and E'

the vertex, P any point on the arc, and A'P'B' the circle

on A'B' as diameter.

Draw PN perpendicular to A'B' meeting the circle in

P' and let the tangent at P meet the tangent at Bin B.

Then, as in Art. 186, PB is parallel to P'B' and the

arc B'P is twice the line B'P'.

Hence ^ PBx = t FB'x = t B'A'P' = 8.

Let a particle start from rest at a point F of the arc

and slide down the arc and let P be its position at anytime.

Its tangential acceleration = g sin d = g sin B'A'P'

HF_g_= 9 = f- . arc B'P, where A'B' is 2a.

Hence the particle moves so that its tangential accele-

ration is always along the arc toward B' and equal to

r/.arc B'P.

288 SIMPLE PENDULUM.

Hence, by Art. 184, the time of a complete oscillation

/4ais 2?r A/ and is therefore independent of the position of

c/

the point F from which the particle started.

Hence the time of oscillation in an inverted cycloid is

independent of the extent of the oscillation.

On account of this property the cycloid is called the

isochronous curve under gravity.

191. The property proved in the previous article will

be still true if, instead of the material curve, we substi-

tute a string tied to the particle in such a way that the

particle describes a cycloid and the string is alwaysnormal to the curve.

This, by Art. 189, may be done by attaching the stringat E' and compelling it to wrap and unwrap itself

alternately upon two metal cheeks in the shape of the

arcs E'QC and E'Q'C' ;in actual practice a pendulum is

only required to swing through a small angle so that only

very small portions of the two arcs near E' are required.

192. Simple Pendulum. A particle tied to a string,

one end of which is fixed, and swinging in a vertical

plane is called a simple pendulum.

Referring to Fig. Art. 189, since the arc E'C is

equal to 4a, a pendulum of length I describes a cycloidthe radius of whose generating circle is a, where I = 4a.

Hence the time of a complete oscillation of the pendu-lum =

periodic time in the cycloid

SMALL OSCILLATION IN A CIRCLE. 289

193. If, instead of compelling the particle in the

simple pendulum to describe a cycloid, we allow it to

describe a circle about the other end as centre the time of

oscillation is still very approximately constant, providedthe angle through which the string oscillates be small.

This can be shewn as follows.

Theorem. If a particle be tied by a string to a fixed

point and allowed to oscillate through a small angle about

the vertical position, to shew that the time of a complete/T

oscillation is 2fr \/-

,where 1 is the length of the string.

Let be the fixed point, OA a vertical line, AP a

portion of the arc described by the particle, and let the

angle AOP be 6.

If PT be the tangent at P meeting OA in T, the

acceleration of the bob along PTL. D. 19

290 SECONDS PENDULUM.

= g sin d

= gQ (approximately, if 6 be small)

=jx arc AP.

.'. as before, the time of a double oscillation is inde-

pendent of the extent of the oscillation and is equal to

27T

194. The above result although not mathematicallyaccurate is very approximately so. If we take into con-

sideration the angle a through which the pendulum swingson each side of the vertical and neglect powers of sin a

above the second, the time of oscillation is

[For a proof the student may refer to Greenhill's Solutions of Cam-

bridge Problems and Riders, 1875.]

If a be 5, the result of the preceding article is within

one-two thousandth part of the accurate result, so that a

pendulum which beats seconds for very small oscillations

would lose about 40 seconds per day if made to vibrate

through 5 on each side of the vertical.

195. Length of a seconds pendulum. A seconds

pendulum is one which makes half a complete oscillation

(viz. from G to C' in Fig. Art. 189) in one second.

Hence, if I be its length, we have 1 = TT A/-

.

\J

_ 9

VARIATIONS IN THE VALUE OF GRAVITY. 291

For an approximate value, putting g = 32'2 and IT =%JL,

we have I = 39'12 inches.

If we use the centimetre-second system we have

/ = 99 3 centimetres.

For the latitude of London more accurate values are

3913929...inches and 99'413...centimetres.

196. The simple pendulum of which we have spokenis idealistic. In practice, a pendulum consists of a wire

whose mass, although small, is not zero and a bob at the

end which is not a particle. Whatever be the shape of

the pendulum, the simple pendulum which oscillates in

the same time as itself is its simple equivalent pendu-lum.

The discussion of the connection between a rigid bodyand its simple equivalent pendulum is not within the

range of this book. We may, however, mention that a

uniform rod, of small section, swings about one end in the

same time as a simple pendulum of two-thirds its length,

whilst the length of the simple equivalent pendulum of a

circular lamina swinging about an axis through the end

of a diameter perpendicular to its plane is three-quarters

of the diameter of the lamina.

197. Acceleration due to gravity. Newton shewed as

a fundamental law of nature that every particle attracts

every other particle with a force which varies directly as

the product of the masses and inversely as the square of the

distance between them. From this fact it can be shewn,

as in any treatise dealing with Attractions, that a sphere

attracts any particle outside itself just as if the whole mass

of the sphere were collected at its centre, and hence that

the acceleration caused by its attraction varies inversely as

192

292 CHANGES IN THE TIME OF

the square of the distance of the particle from its centre;

similarly the attraction on a particle inside the earth can

be shewn to vary directly as its distance from the centre

of the earth.

Hence if g1be the value of gravity at a height h above

the earth's surface, g the value at the surface, and r the

earth's radius, then gl: g :: .

2 : -^ ,

so that gl=

So if <72be the value at the bottom of a mine, of depth

j r- da, we have gz

= g-

.

198. We will now investigate the effect on the time of oscillation of

a simple pendulum due to a small change in the value of g, and also the

effect due to a small change in its length.

If a pendulum, of length 1, make n complete oscillations in a given time,

to shew that

(1) If g be changed to g + G, the number of oscillations gained

. n Gw )

2 g'

(2) If the pendulum be taken to a height h above the earth's surface

the number of oscillations lost is n -, where r is the radius of the

earth,

(3) If it be taken to the bottom of a mine of depth d, the number lost is

nd2 r

'

(4) If its length be changed to 1 + L, the number lost is -.

'2 L

Let T be the original time of oscillation, T' the new time of oscilla-

tion, and n' the new number of oscillations in the given time so that

nT=n'T'.

OSCILLATION Of A PENDULUM. 293

(1) In this case T=2ir Jlfg; T'=2*Jll(g + G).

n' T /. G 1 GHence - = -=*/ 1 + - ! + 5

~' approximately,

( using the Binomial Theorem and neglecting squares of -j

.

Hence the number of oscillations gained = n' - n= s .

*i Q

So if g become g -G, the number lost is -* 9

(2) If g - G be the value of gravity at a height h we have

9-G_ i* _/V, ^-2_, 2ft

9

.. G=y and hence, as in (1), the number of oscillations lost is

hn -

.

r

(3) If g - G be the value at a depth d we have g-G : g :: r-d : r,

1 /*"* /?

hat the number of oscillations lost = H = - -.

2<; 2r

(4) when the length I of the pendulum is changed to Z + L we have

L\-i . IL

Hence the number of oscillations lost= n - ri = - .

/

199. From the preceding article it follows that the

height of a mountain or the depth of a mine could be

found by finding the number of oscillations lost by a

pendulum which beats seconds on the surface of the

earth.

If the point above the sea level at which the pendulumbeats be on a considerable table land the effect of the

attraction of the portion of the earth which is between

294 FINDING "g" BY MEANS OF THE PENDULUM.

the pendulum and sea level could be allowed for by

taking the value of gravity there as g (1

)instead of

\ 4/*/

l + -) and then the number of oscillations lost wouldrj

, 5n h . , f hbe - - instead of n -

.

8 r r

200. The value of g has been most accurately found

by means of timing the oscillation of a pendulum. This

may be done as follows.

The pendulum on which we experiment is set swinging

in front of a clock whose pendulum we know to be beating

true seconds, and they are started so that they are both

vertical at the same instant and the time is carefully

noted at which they are again both vertical together and

moving in the same direction; we then know that the

pendulum has gained (or lost) one complete oscillation on

the seconds pendulum. Suppose, for example, that they

coincide at the end of 300 seconds, then in that time the

seconds pendulum has made 150 complete oscillations and

the other pendulum 149 so that the time of its complete

oscillation would be ff seconds.

The time of an oscillation may be found much more

accurately by finding the time that would elapse before

the experimental pendulum had lost a large number, say

20, oscillations and then taking the average.

The length of the simple equivalent pendulum corre-

sponding to our experimental pendulum may be then

measured and the value of g determined by means of the

formula T=2

EXAMPLES ON CHAPTER VIII. 295

201. Verification of the law of gravity by mean* of the moon'$

nidtion. We may show roughly the truth of the law of gravitation by

finding the time that the moon would take to travel round the earth on

the assumption that it is kept in its orbit by means of the earth's attrac-

tion.

Let / be the acceleration of the moon due to the earth's attraction ;

then, since the distance between the centres of the two bodies is roughly

60 times the earth's radius, we have/ : g :: .^

: -%,so that/= 0/3600.

Let r be the velocity of the moon round the earth so that

v2/60r=/ =0/3600.

.=-.60

.. the periodic time of the moon = 2ir x 60r ~ v = 2ir . 60 x . / seconds.

Taking the radius of the earth to be 4000 miles and g as 32-2 this time

is 27 g 4 days, which is approximately the observed time of revolution.

EXAMPLES. CHAPTER VIII.

1. How many oscillations will a pendulum of length

53'41 centimetres make in 242 seconds at a place where

<7=981?

2. Shew that a pendulum one mile in length will oscillate

22 minutes nearly.

3. A pendulum 37 '8 inches long makes 183 beats in

three minutes at a certain place ;find the acceleration due to

gravity there.

4. A pendulum oscillating seconds at one place is carried

to another place where it loses two seconds per day ; compare

the accelerations due to gravity at the two places.

296 EXAMPLES ON CHAPTER VIII.

5. A clock which at the surface of the earth gains 10

seconds a day loses 10 seconds a day when taken down a

mine; compare the accelerations due to gravity at the topand bottom of the mine and find the depth of the mine.

6. If a seconds pendulum be carried to the top of a

mountain half a mile high how many seconds will it lose per

day, assuming the earth's centre to be 4000 miles from the

foot of the mountain, and by how much must it be shortened

so that it may beat seconds at the summit of the mountain 1

7. Shew that the height of a hill at the summit of which

a seconds pendulum loses n beats in 24 hours is approximately245 . n feet.

8. If a pendulum oscillating seconds be lengthened by its

hundredth part, find the number of oscillations it will lose in

a day.

9. A simple pendulum performs 21 complete vibrations

in 44 seconds;on shortening its length by 47*6875 centimetres

it performs 21 complete vibrations in 33 seconds; find the

value of g.

10. A clock with a seconds pendulum loses 9 seconds

per day ;find roughly the required alteration in the length of

the pendulum.

11. If ^ be the length of an imperfectly adjusted seconds

pendulum which gains n seconds in an hour and 12the length

of one which loses n seconds at the same place ;shew that the

true length of the seconds pendulum is the harmonic mean

between ^ and 12.

12. A balloon ascends with a constant acceleration and

reaches a height of 900 feet in one minute. Shew that a

pendulum clock carried in the balloon will gain at the rate of

27 '8 seconds per hour.


13. Two clocks identical in all respects are placed at

opposite points on the earth's surface, at one of which the

moon is vertically overhead. Assuming the earth and moonto remain at rest and that the mass of the earth is 80 times

that of the moon, shew that if the clocks be started together

they will at the end of twenty-four hours differ by about f^thsof a second.

14. The bob of a pendulum which is hung close to the

face of a vertical cliff is attracted by the cliff with a force

which would produce an acceleration f in the bob; shew that

the time of a complete oscillation is 2ir,JP/(g2

+fy),

where I

is the length of the pendulum.

15. Prove that a seconds pendulum, if carried to the

moon, would oscillate in If seconds, assuming the mass of the

earth to be 49 times that of the moon and the radius of the

earth 4 times that of the moon.

16. A railway train is moving uniformly along a curve

at the rate of 60 miles per hour and in one of the carriages a

seconds pendulum is observed to beat 121 times in two

minutes. Prove that the radius of the curve is about a

quarter of a mile.

17. Assuming the earth to be a sphere whose attraction

per unit of mass is the same at all points of its surface, provethat a pendulum which beats seconds at the poles will lose

30 ra cos2 A beats per minute in latitude X, where the ratio of

the weight of the body at the poles to its weight at the

equator is 1 + m to 1.

18. If a seconds pendulum be formed of a heavy particle

suspended by a string of length I from a point A in a vertical

wall, and if a nail jut out from the wall at a distance -n

below A so as to catch the string once in each completeoscillation find the time of oscillation, n being large.

298 EXAMPLES ON CHAPTER VIII.

1 9. In cycloidal motion the vertical velocity of the particle

is greatest when it has described half its vertical descent.

20. A particle of mass m falls down a cycloid under

gravity starting from the cusp ;shew that the pressure on the

cycloid at any point is 2mg cosif/,

whereif/

is the inclination

to the horizon of the tangent to the cycloid at that point.

21. When a particle falls from rest down the arc of a

cycloid prove that it describes half the path to the lowest

point in two-thirds the time to the lowest point.

22. If a particle slide down a smooth cycloid, whose axis

is vertical and vertex downwards, from a point whose arcual

distance from the vertex is b, prove that its velocity at time t

from the start is sin,where T is the time of a complete

oscillation of the particle.

23. A particle is just displaced from rest at the vertex of

a smooth cycloid, whose axis is vertical and vertex upwards,

and which rests on a horizontal plane ;shew that it will leave

the cycloid when its direction of motion makes an angle of 45

with the vertical and that it will strike the plane at a time

-1) / - after it leaves the cycloid, a being the radius of

the generating circle of the cycloid.

24. Shew that the time a train, if unresisted, takes to

pass through a tunnel under a river in the form of an arc of

an inverted cycloid of length 2s and height h cut off bya horizontal line is

where v is the velocity with which the train enters and leaves

the tunnel.


25. A particle moves in a circle drawn on the deck of a

ship with a uniform speed equal to that of the ship. Find

the path it describes in space. Also find the path of the

particle if the given uniform speed be less than that of

the ship.

26. A pendulum, of length I, has one end of its string

fastened to a peg on a smooth plane inclined at an angle a to

the horizon;with the string and the weight on the plane its

time of oscillation is two seconds. Find a having given that

a pendulum of length O -T^ oscillates in one second when

suspended vertically.

27. In a conical pendulum shew that the time of revolu-

tion is ultimately 2ir I - when the inclination is very small.V 9

28. If the length I of a simple pendulum be comparablewith the radius a of the earth, shew that its time of oscillation

is Trjla/g (a I).

29. A point moves in a path produced by the combination

of two simple harmonic motions of equal amplitude in two

rectilinear directions, the periods of the components being as

1:3; find the paths described when the epochs are the same

and when they differ by 90.

30. If a frictionless airless tunnel were made from Londonto Paris, prove that a train would traverse the tunnel under

the action of gravity in less than three-quarters of an hour.

Shew that the same theorem would be true whatever two

points were taken on the earth's surface.

CHAPTER IX.

MOTION OF A PARTICLE ABOUT A FIXED CENTRE

OF FORCE.

202. Moment of a velocity. Def. The moment

of the velocity of a moving point about a fixed point is the

product of the velocity of the moving point and the perpen-dicular drawn from the fixed point upon the direction ofmotion of the moving point.

If be the fixed point and PA represent in magni-tude and direction the velocity of the moving point at a

point P of its path, the moment of its velocity about is

MM.Ml-:NT <>1 A VKUM'ITY. 301

represented by PA . Oy, where Oy is the perpendicular

upon PA. Hence this moment may be represented

geometrically by twice the area of the triangle OPA.

203. Theorem. If a point describe a curve in one

plane, and its acceleration be always directed towards a

fixed point in the plane, the moment of its velocity about the

fixed point is unaltered.

Let PQ be a portion of the path of the moving point,

and let the tangents at P and Q meet in T. Let TP',

TQ' represent the velocities at P and Q.

Let be the fixed point ; join OT and P'Q'. Since,

whilst the point describes the arc PQ, its velocity is

changed from TP' to TQ', the total change of velocity is

represented by some line passing through T. But, since

the acceleration of the point is always directed towards

0, the direction of this change of velocity must pass

through 0. Hence the change of velocity whilst the

302 MOTION IN AN ELLIPSE WITH

moving point describes the arc PQ is in the direction

TO.

But, as in Art. 24, this change of velocity is repre-sented by P'Q'.

.-. P'Q' is parallel to TO.

Now the moments of the velocities at P and Q about

are represented by twice the areas of the triangles

OTP, OTQ' respectively.

Also, since TO and P'Q' are parallel, these two tri-

angles are equal.

Hence the moments of the velocities at P and Q about

are the same.

This is true whatever points P and Q are taken on

the path.

Hence the theorem is proved.

Cor. By Art. 40, the areal velocity of the moving

point about is J v .p and it is therefore constant.

Hence the area traced out by the line joining the

moving point to is proportional to the time of describingthe area, i. e. Equal areas are described in equaltimes. This is the form in which Newton enunciates

the proposition in Book I., Section II., Prop. I. of the

Principia.

204. Theorem. A particle is projectedfrom a given

point, with any velocity and direction, and moves with an

acceleration which is always directed towards a fixed pointand varies as the distance from the fixed point ; to shew

that its path is an ellipse whose centre is at the fixed point,

and to determine the other circumstances of the motion.

Let C be the fixed point towards which the accelera-

AN ACCELERATION DIRECTED ToWAUDS THE CENTRE. 303

tii in is always directed, A the point of projection, and Vthe velocity of projection.

Draw CB parallel to the direction of projection. Let

P be the position of the particle at the end of time t and

draw PN, PM parallel to CB and CA to meet CA and

CB in N, M respectively. Let the acceleration towards

C befj.

. PC. By the triangle of accelerations this accele-

ration is equivalent to accelerations represented by p, . PNand p . NC, i.e. by /* . MC and /* . NC.

Hence the points N, M move in the lines AC, CB with

simple harmonic motion.

Make CB equal to F2

//*; then, by Art. 178 (1), CB is

the amplitude of the harmonic motion of the point M.

Also CA is the amplitude of the motion of the

point N.

Let CA, CB be a and b' respectively.

By Art. 178, (2) we have CN = a' cos >Jpt.

7TAlso since t is the time from C to M and since

^-T-

7Tis the time from C to B, therefore ^-t t is the time

from M to B, or from B to M.

304 PERIODIC TIME AND VELOCITY AT ANY POINT.

,'. PN=V cos -/itt

= b' sin

Hence _ _.

Hence P moves on an ellipse of which CA and CB are

a pair of conjugate diameters.

Periodic Time. The time of describing the ellipse

is equal to the periodic time in the simple harmonic

motion, and hence is -7- .

V>If h be twice the area described in the unit of time, we

have

zr x -r- = area of the ellipse = Trab,2 VA1

where 2a, 26 are the major and minor axes of the ellipse.

Hence h = J/j, ab.

Velocity at any point. If v be the velocity of the

particle at P, and CF be the perpendicular from G uponthe tangent at P, we have

h _ ///, ab=CF

==

~CF~'

But if CD be the semi-diameter conjugate to CP, we

have CF . CD = ab.

.'. v = <Jp, . CD, and hence the velocity at any point Pvaries as the diameter conjugate to CP.

MOTION IN A CONIC SECTION. 305

205. Theorem. A particle is moving in an ellipse

with an acceleration which is always directed towards one

of the foci ; to shew that the acceleration varies inversely

as the square of the focal distance and to find the other

circumstances of the motion.

Let S be the focus towards which the acceleration is

always directed, H the other focus, and P the position of

the particle at any time. Let v be the velocity of the

particle at P.

Draw SY, HZ the perpendiculars upon the tangentat P.

Since the acceleration is always towards S,

.-. by Art. 203, we have

.A..AiM - h HZ~SY SY.HZ'V"if 6 be the semi-minor axis, and h be twice the area

described in unit time.

L. D. 20

306 MOTION IN A CONIC SECTION WITH

Hence HZ is perpendicular to and proportional to the

velocity ;therefore the hodograph of the motion is the

locus of Z [i.e. the auxiliary circle] turned through a right

angle about H.

Hence the acceleration in the path is perpendicularand proportional to the velocity of the point Z.

Now, since CZ is parallel to SP, the angular velocity

of Z about C is equal to the angular velocity, w, of the

particle about 8, and hence the velocity of Z is aw, where

a is the semi-major axis.

/. the acceleration of P = T, . aw.b2

But, by Art. 40, a> = -7 ,where SP is r,

T

:. the acceleration of P = --,-

b r

Hence the acceleration of the particle is -3 ,

where

-rj=

/*, and therefore

h = V/A x setni-latus-rectum.

Periodic Time. The periodic time

-j- A / a- =V a

Velocity at any point. If 8Y, HZ be p, p' respec-

tively, we have v = -.

But the triangles SPY, HPZ are similar, and there-

AN ACCELERATION DIRECTED TOWARD THE FOCUS. 307

l>

t

= ptf = V^

T TT TT

=//' r r

i_h*__h* 2a - r__ p 2a-r _ /2 1

.'. 17 ..j ., . i4 I

p b r a r \r a

Cor. 1. If in the previous proposition we make a

b3

infinite whilst the serai-latus-rectum,-

,remains finite, the

CL

ellipse becomes a parabola, and the expression for the

2it

square of the velocity at any point becomes .

Gor. 2. If we apply the reasoning of the proposition

to the case of a hyperbola we should find that a particle

would describe a branch with an acceleration ~ directed tor

1

the focus belonging to the branch;in this case, since in

the hyperbola the difference of the focal distances of any

point is equal to the major axis, the square of the velocity

/2 1\would be u, - + -

.

\r a)

206. Conversely, if a particle be projected from any

point P with any velocity in any direction, and move with

an acceleration which is always directed towards a fixed

point S and is equal to ~T.- -

t ,it will describe a conic

(distance)2

section having S as one of its foci;also the path will be

an ellipse, a parabola, or an hyperbola, according as the

square of the velocity of projection is less than, equal to,

2aor greater than the quantity ^^ .

202

308 KEPLER'S LAWS.

The orbit may be easily constructed. For the major

/ 2 1\axis of the orbit is given by v* = a

(~

) ,where v is

\8P aj

the velocity of projection. Also the second focus H is

such that SP, HP make equal angles with the direction

of projection and such that SP + HP is equal to the

major axis. Hence the position of the major axis is de-

termined.

207. The points A and A', Fig. Art. 205, at which

the particle is moving in a direction perpendicular to the

line joining it to the centre of acceleration, are called the

Apses of the orbit. The line joining them, which is the

major axis of the orbit, is called the Apse-Line.

208. Kepler's Laws. The astronomer Kepler, bymuch patient labour, discovered three laws connecting the

motions of the various planets about the sun. They

are;

1. Each planet describes an ellipse having the sun in

one of its foci.

2. The areas described by the radii drawn from the

planet to the sun are, in the same orbit, proportional to

the times of describing them.

3. The squares of the periodic times of the various

planets are proportional to the cubes of the major axes of

their orbits.

From the second law we conclude, by Art. 203, that

the acceleration of each planet (and therefore the force

acting on it) is directed towards the sun.

EFFECT OF DISTURBING FORCES. 309

From the first law we conclude, by Art. 205, that the

acceleration of each planet varies inversely as the squareof its distance from the sun's centre.

From the third law we conclude, by Art. 205, that the

absolute acceleration/j, [that is, the acceleration at unit

distance from the sun] is the same for all the planets.

209. Effect of disturbing forces on the path of a particle. Tangential

disturbing force.

Let APA' be the path of a particle moving about a centre of force at

S and let H be the other focus.

When the particle arrives at P let its velocity be changed from v to

v + u, the direction being unaltered ; let 2a' be the new major axis.

Then we have

.'. by subtraction, (v + u)z-v^=fj. (

- -, ),

giving the change in the major axis.

Since the direction of motion at P is unaltered the new focus lies on

PH ; and, if H' be its position, we have

H'H= (H'P + SP)- (HP+ SP) = 2a' - 2a.

Hence we have the direction of the new major axis.

210. If the disturbing force be not tangential, the velocity it produces

must be compounded with the velocity in the orbit to give the new

velocity and direction of motion at the point P. This direction will be

the new tangent at P;as in the last article we can* now determine the

310 EFFECT OF A CHANGE IN ABSOLUTE ACCELERATION.

new major axis. Also from the fact that the moment of the velocity of

the point about the focus is equal to J/j. x semi-latus-rectum we obtain

the new eccentricity. Finally, by drawing a line making with the new

tangent at P an angle equal to that made by SP, and taking on it a point

H', such that SP + H'P is equal to the new major axis, we obtain the newsecond focus, and hence the position of the major axis of the neworbit.

211. Effect on the orbit of a change in the value of p. When the

particle is at a distance r from the centre of force let the value of p. be

changed toyoi', and let the new values of the axis-major and eccentricity

be 2A and E.

Since the velocity is instantaneously unaltered in magnitude wehave

'2 1\ , /2 r

which is an equation to determine A.

The moment of the velocity about S being unaltered, h remains the

same.

.-. JIM (1- e

2)=

\llJ-'A (1- .E8),

giving E.

The direction of the velocity at distance r being unaltered we obtain

the new position of the second focus, and hence the direction of the new

major axis as in Art. 209.

212. Motion in a straight line with an acceleration

always directed toward a faced point in the straight line

and varying inversely as the square of the distance.

The velocity when the point has described any distance

from rest can be easily deduced from the results of

Art. 205.

For the velocity at any point P of the ellipse is given

2

Let SA be c so that c = a (1 + e).

2 1+e

MOTION IN A STRAIGHT LINE. flll

Let the minor axis of the ellipse be now indefinitely

diminished, the major axis being unaltered in length ;

since 6* = a8

(1 e8

), the eccentricity e will become unity.

Also the ellipse will degenerate into the straight line SA,and P will become a point on SA.

The expression for the velocity will become

2 2

where SA = c.

Hence, if a particle move in the straight line AS under

an acceleration -.,. ^ r= towards S, its velocity when at(distance)

a distance a; from S will be given by

Also the time from A to S

= Lt.fl=1 (Time of describing one half of the ellipse)

213. Motion of a projectile, taking into consideration

variations ofgravity.

We have pointed out, Art. 197, that the attraction of

the earth on a point outside the earth at a distance r from

312 MOTION OF A PROJECTILE, VARIATIONS

its centre is^. Hence the motion of a projectile, the

resistance of the air being neglected, is a particular case

of Art. 205, one of the foci of the ellipse described by the

particle being at the centre of the earth.

It will be noted that the value of gravity at the

surface of the earth is ~,where R is the radius of the

earth. Hence-j^=

g, so that//,= gR*.

In the next article will be discussed a few elementarytheorems on this portion of the subject.

214. Let 8 be the centre of the earth and P the

point of projection. Let K be the point vertically above

IN GRAVITY BEING CONSIDERED. 313

P to which the velocity, V, of projection is due, so that,

by Art. 212, we have

where R is the radius of the earth and PK is h.

If H be the second focus of the path the semi-major

axis is (R + PH}. Hence, by Art. 205,

By comparing this with equation (1) we have PH = h,

so that the locus of the second focus is, for a constant

velocity of projection, a circle whose centre is P and

radius h. It follows that the major axis of the path is

SP+PHorSK.The ellipse, whose foci are 8 and H, meets a plane

LPM, passing through the point of projection, in a point

Q such that SQ + QH= SK. Hence, if SQ meet in T the

circle whose centre is S and radius SK, we have QT= QH.Since there is, in general, another point, H', on the circle

of foci equidistant with H from Q we have in general two

paths for a given range.

The greatest range on the plane LPM is clearly Pqwhere qt equals qO.

Hence _Sq + qP = Sq + qO + OP = Sq + qt +PK= SK+PK.

Therefore q lies on an ellipse whose foci are the centre

of the earth and the point of projection and which passes

through K.

Hence we obtain the maximum range.



1. Find the velocity acquired in falling through a height h to the

earth's surface, taking into consideration variations in the acceleration due

to gravity.

If R be the radius of the earth the acceleration at a point outside the

earth is .-=-.-

, where = q.(distance)

2 R-

Hence the velocity of the particle on reaching the earth is given by

R+ hJ' R + h'

Hence the square of the velocity acquired is less than if gravity were

constant by 2gh-^~-,that is, by 2gh*l(R + h).

2. If the earth were suddenly stopped in its orbit find the time tJiat

would elapse before it fell into the sun, neglecting the eccentricity of its

orbit.

Let R be the radius of the orbit of the earth and w its angular velocity

about the sun.

The attraction of the sun is /u/(distance)2

, where-^r,=w

5^, so that

M= U3w2.

By Art. 212, the time required

*-=-^2x/2w 4w

'

f\

But = period of revolution of the earth 365 days,w

.'. time required= ~ days = about 64 days,o

If we do not neglect the eccentricity, e, of the orbit the time would lie

8

between 64J x (le)* days.

3. A small meteor, of mass m, falls into the sun when the earth is at

the end of the minor axis of its orbit; ifM.be the mass of the sun, find the

changes in the major axis, in the position of the apse line, and in the

periodic time of the earth.


A'f -i- tn

The new value of M is n ,- Hence, if 2A be the new major axis,.17

we have/2 1\ M+mf2 1

M - - -)=

/* Tf

so that

If .S' be the sun, U the original second focus, and H' the new second

focus, we have HH'= (SP+PH)-(SP+ PH') = 2a-2A

m b

rr/orr_ HH' BW H'HS "jj'a b~

2ae - 2a . e

giving the angle through which the major axis is turned.

1-rr &Also the new periodic time=

so thai the periodic time is shortened by of a year.JE

EXAMPLES. CHAPTER IX.

1. A particle describes an ellipse under a force to its

centre;shew that its angular velocity about a focus is in-

versely proportional to its distance from that focus.

2. A particle is describing an ellipse under a force to the

centre;

ift',

vl ,

v2 be the velocities at the extremities of the

latus rectum and major and minor axis respectively, prove that

316 EXAMPLES ON CHAPTER IX.

3. A particle describes an ellipse about a centre of force

in the focus;shew that the velocity at the mean distance

from the centre is a mean proportional between the velocities

at the ends of any diameter.

4. If a particle describe an ellipse about a centre of force

in one of the foci, shew that its velocity at any point of its

path may be resolved into two constant velocities respectively

perpendicular to the major axis and the radius vector to the

centre of force.

5. A particle describes an ellipse about a centre of force

in the focus;shew that the angular velocity about the other

focus varies inversely as the square of the normal at the

point.

6. Shew that if the velocity of the earth at any point of

its orbit were increased by about one-half it would describe a

parabola about the sun as focus.

7. If a body be projected from the earth with a velocity

exceeding 7 miles per second it will not return to the earth,

and may even leave the solar system.

8. If the perihelion distance of the parabolic orbit of a

comet be equal to the mean distance of the earth from the sun,

find the number of days the comet will take from one end of

its latus rectum to the other.

9. Given that there are 13 lunar months in a year, that

the sun's angular radius as seen from the earth is 16', and

that the earth's angular radius as seen from the moon is one

degree, shew that the mean density of the sun is about one-

third that of the earth.

10. Taking into consideration the variations in the value

of gravity outside the earth's surface, shew that the maximum

EXAMPLES ON CHAPTER IX. 317

range of a projectile on a horizontal plane passing through the

point of projection is tgu?!? / (ig'J?-

u'), where u is the

velocity of projection and Ji is the radius of the earth.

11. Shew that a gun at the sea-level can command - ofn

the earth's surface if the greatest height to which it can send

the ball be - th of the earth's radius, and that the requiredn

In 1direction of projection is inclined at an angle tan" 1

* / to^J 72- T~ L

the horizontal, variations of gravity being taken into account.

12. If the velocity of projection from a point on the

surface of the earth be such that directed vertically upward it

would carry a body to a height equal to the earth's radius,

prove that, if the direction of projection make an angle of n

with the vertical, the range on the earth's surface will be

120w nautical miles.

13. A particle is describing an ellipse under a force to

the focus;when it comes to the end of the minor axis the

absolute force is diminished by one-third. Find the position

and dimensions of the new orbit, and shew that the distance

between its focus and its centre is bisected by the minor axis

of the original orbit.

14. A particle is describing an ellipse under the action of

a force to the focus, and when it arrives at the end of the minor

axis the magnitude of the force is reduced to one-third of its

original value; shew that the orbit will become hyperbolic

and that its conjugate axis will be to the minor axis of the

ellipse in the ratio of ^/S : 1.

15. When a particle is at the nearer apse of an ellipse

described about the focus the absolute force is increased by a


small fraction, -th, of itself: when at the further apse then

absolute force becomes less than the original value by this

amount. Shew that the time taken in this revolution is equalQe

to the original period reduced by ; of itself.n (1

- e2

)

16. If when a particle is at any point P of an elliptic

orbit the centre of force be suddenly moved from the focus Sto the other focus H, prove that the intensity must be altered

in the ratio HP2: SF* so that the particle may continue to

describe the same path.

17. A body is revolving in an ellipse under the action of

a force to the focus;on arriving at the end of the minor axis

the law of force is changed to that of the direct distance, the

magnitude of the force at that point remaining the same;

shew that the periodic time will be unaltered and that the

sum of the new axes is to their difference as the sum of the

old axes to the distance between the foci.

18. If when a particle is at the extremity of the minor

axis of an ellipse which it is describing about the focus, the

centre of force be suddenly removed a small distance aa

toward the particle, the eccentricity will be unaltered but the

major axis will be turned through an angle whose circular

measure is a [e~2

1] .

19. A planet, of mass M and periodic time T, when at

its greatest distance from the sun comes into collision with a

meteor, of mass m, moving in the same orbit in the oppositefVYl

direction with velocity v',

shew that, if -^ be small, the major

axis of the planet's path is reduced by

4m vT l\ -e~M

'

~^ v r+'

EXAMPLES ON CHAPTER IX. 319

20. A planet, when at the extremity of the latus-rectum

of its path, meets with a shower of meteors which reduces

its velocity by -th without altering the direction; prove7%

3 1 + e9

that the periodic time is reduced by-

^ of itself.71 1 ^ 6/

21. The velocity of a body moving in an ellipse about the

focus is increased in the ratio l+n : 1, where n is small;

shew that the corresponding increase in the eccentricity is

2n (e + cos 0) where is the angle ASP, P being the position

of the body, S the focus, and A the apse.

Shew also that the apse-line is turned through an angle

2n sin 6

e

22. A particle describing an ellipse about a centre of force

in the focus is checked, when at P, by a small impulse yT8

along the tangent at P, where V is the velocity at P; prove

that the major axis is turned through an angle-* sin PSH

and find the change in the eccentricity.

23. A particle is revolving in an ellipse about its focus

and is at the end of the minor axis and moving away from the

centre of force when it receives a blow which alters its path

into a circle. Shew that the impulse of the blow is equal to

2u .- and find also the direction of the blow.aa

24. A body is describing an ellipse about a centre of

force in the focus and when its radius vector is one-half the

latus rectum it receives a blow which causes it to move toward

the other focus with a momentum equal to that of the blow;

find the position of the axis of the new orbit and shew that its

1 e2

eccentricity is^ ,

where e is the eccentricity of the original

orbit.


25. A body is moving in an ellipse about a centre of force

in the focus; when it arrives at P the direction of motion is

turned through a right angle, the speed being unaltered;shew

that the body will describe an ellipse whose eccentricity varies

as the distance of P from the centre.

26. If a body (moving in an ellipse under a force to

the focus) when at the mean distance have the direction

of its velocity turned through an angle a, the eccentricity of

the new orbit will be e cos a /.fT e2 sin a, where e is the

original eccentricity.

If a be small and at the same time the velocity be altered

by e -~-a, shew that the apse-line is unaltered.

27. A body, of mass in, describes an ellipse under a force

to the focus whose absolute magnitude is/JL.

When the body

is at the end of the minor axis it receives a blow in the

direction of the normal of magnitude m /-, the intensity

of the force being at the same time doubled;shew that the

magnitude of the axis of the new orbit is unaltered in magni-

tude, and that the new eccentricity is -j (e + Jl - es

).If the

v^intensity had remained the same, shew that the new orbit

would have been a parabola.

28. A body is moving in an ellipse about a force in the

focus;when at the end of the latus rectum through the centre

of force it receives a blow which makes it move at right angles

to its former direction;shew that the new orbit will be a

hyperbola concentric with the ellipse if the ratio of the

velocities before and after the blow be e : 1.

ANSWERS TO THE EXAMPLES.

CHAPTER I. (Pages 5055.)

a - >J<)h ; zero and <Jgh, where h is the height of the plane.

16000 IT8 -

"Si ' 5 "

. |f ft.-sec. units; f ft. -sec. units; 40 miles per hour.

1O. 60 miles per hour; 44 sees.; 1936 feet; 8 sees.

14. If P be the point of the circle furthest from the given straight

line, and Q be the point moving in the circle, the velocity of Q with

respect to the other is PQ . u perpendicular to QP.

3O. w A/ , where w is the angular velocity of the earth about its

axis.

35. (1) The chord makes an angle of 60 with the vertical; (2) the

chord makes an angle sec" 1^/3 with the vertical.

37. Through the end of the latus rectum and inclined at 60 to the

horizon respectively.

30. The length of the chord is b>J2.

37. A circle passing through the lowest point of the given circle and

touching it there.

L. D. 21

322 ANSWERS TO THE EXAMPLES.

CHAPTER II. (Pages 103113.)

1. | > 5 ,and poundals respectively.

u O 1

4. Weight of 9f| tons. 6. 54,650,956,800 absolute units.

7. 20^2 feet per second; 560,000 foot-pounds.

8. 68AVH.-P.

10. Weight of llfi tons; 28,233,333^ foot-pounds.

11. About 50-4 and 23-1 feet per second respectively.

12. -938 tons weight. 13. 1260 r\ poundals.

14. ISO^Vs- 15 - 7270. 16. 11T\ tons.

17. 1 minute 42 seconds.

19. The parliamentary train by 9'46 per cent.

21. 9T9T miles per hour.

23. The straight line which bisects the distances between the eaves

and the ridge and ground respectively.

24. ^/6 feet per second ; | ^30 feet per second.

27. 8mm'l(m + m') units of impulse ; 8m/(m + m') feet per second.

M-M'29. After M reaches the ground, M' describes a distance h

,on

the plane and then returns.

30. 660,000 foot-lbs. ; 30 H.-P. 32. -169 : 1.

36. '88 foot-tons. 41. 200 feet above;1000 feet below.

43. -,where M is the common velocity.

45. i -ft

2,where / is the acceleration produced by the force

n

52. (P- W)g/W; the tension at any point distant x from the lowest

point is - P, where a is the whole length of the chain.

ANSWERS TO THE EXAMPLES. 323

CHAPTER III. (Pages 129-142.)

1. The acceleration becomes (M - 8m+ 7n) gl(M + Mm + 21/u).

TO,-

2/H., 7W,- 2/Ko \\in.vi.,

6. 0; 2 -;/: g poundals; it is assumedr! + 4 2 TOJ + 4TO.J n^ + 4m2

that all the strings are vertical.

o. If M be fixed its acceleration is(M -m- m') gj(M + m + m') .

13. 4ww'/(8m - m').

14. W1will remain at rest if tWl

= W3+ 4Wa WJ(Wa + W4).

17. The velocity of the weight struck is five times that of each of the

others and is in the opposite direction.

19. 1005-2 feet per second.

ao. About 3580 and 3100 feet per second respectively.

as. The ratio is M : M+ m, where M is the mass of the bucket andm is the mass of the frog.

Pm-jn' u a (a + 6) u b(a + b)'

m m+ m''

as.aTn'oTs"

W^ere m IS tne mass of each particle, u the velocity

with which C is projected, and a the angle ABC.

30. 57fH.-P. 81. 2^H.-P.33. 125ir/96 revolutions

; 125w/24 revolutions.

34. The ring comes to rest when its length is 2irx where

W [Jr^-a*- Jr* - a;2] a= irX (x

-o)

2,

W being the weight of the ring, r the radius of the sphere, 2ra the initial

length and X the coefficient of elasticity of the ring.

CHAPTER IV. (Pages 164166.)

1. iju- feet ; a seconds ;- Ibs.a

a. 4 -3...grammes; 18-21...metres; 5*45 seconds.

3. 2400 feet ; 15 seconds ; 600 Ibs.

4. 8800 yards ;5 minutes ;

54,", tons.

6. 1 mile ; 8 minutes ; 09ff tons.

0. 3: 1; 9: 1; 2 : 15. 7. [T]2[J/][]-*.


8.,y/ 2

= about 31300 Ibs.

10. About 33 x 107IT Ibs.

; about 3-35 x 10- 12lbs. weight.

11. [L? [IT* [#]-'. 12. 6-488 x 10-4 dynes.13. 3-98 xlO20

.

CHAPTER V. (Pages 191-201.)

6. A circle of about 91 miles radius. 8. 2J Ibs.

g -2wbere is the velocity of the centre of the wheel and a

is its radius.

12. Its intersection with the horizontal moves in the same manner as

a point which starts with velocity ,J3u and moves with acceleration

3O. In each case the maximum range will be found by drawing the

enveloping parabola ;the angles of projection are tan"1 * /_3-j and

\r fl -j- K

tan"1 A/ -. T respectively, where k is the height of the cliff and h the

height to which the velocity of projection is due.

34. Let A and B be the two given points, B being the lower; draw

BK vertical and AB horizontal to meet it in K; the focus of the required

path is a point, S, on AB such that 2AS=AB-BK; the required point,

C, on the horizontal plane is such that 2CS . AB =AK 2.

36. 6-1 inches. 39. 1 '92 Ibs.

42. The latus rectum is unaltered ; also, if S, S' be the old and newfoci and P be the point of contact, then S, S', and P are in a straight

line, and S'P : SP :: (m-m')2

: (m + m'f, where m and m' are the masses

of the particles.

CHAPTER VI. (Pages 227240.)

6. The vertex of the new path is in the horizontal line through the

starting point at a distance ; the focus is at a depth ^.

22. The condition is that the equation 1 -2en+ e*n+r=Q should give

an integral value of n for an integral value of r.

ANSWERS TO THE EXAMPLES. 325

43. The velocity of each is ultimately mv/(Jl/ + m).

43. 12| feet; if the height be less than this, and the train be long

enough, the ball will rebound from a point of the train behind the point

that it first hits.

1 - e44. If tan a be >

, cot X the horizontal velocity will be destroyed1 + e

before the vertical velocity and the particle will rebound vertically.

CHAPTER VII. (Pages 268 276.)

4. 30,000 poundals.

5.[

- -jpoundals per unit of length, X being the coefficient of

elasticity, J/ the mass of the band and r its radius.

7. 24-31 hours roughly, taking = 32 and *= y.

11. Jag sec o ^3 + 2 sin a - 4 sin3 a.

13. Its mass must be diminished in the ratio 1 : 4.

14.

e36. rcosec-= constant.

i

37. The lesser ball remains at the bottom of the tube and the greaterrises through a height equal to one-half the radius of the tube.

3O. 2 cos-' ^- ; with a load M' the angle is 2 cos" 1 4r -'.. , wheretor M /or*

/ is the length of a rod and u the angular velocity of the system.

36. 18n-a/u seconds;m 2

/a poundals ;mu2

/6a poundals.

37. 2 sjlga.

39. If p be > 4 the particle A will always be at rest.

CHAPTER VIII. (Pages 295299.)

[IT has been taken to be ^ and g to be 32.]

1. 165 complete oscillations nearly. 3. 32-15.

4. 1 : -999953. 5. 1-0046 : 1 ; the depth is 1-85 miles.

6. 10-8 seconds ; -0097 inch. 8. 432. 9. 981.


1O. -0081 inch. 18.

25. A cycloid ; a prolate cycloid. 26. 45.

29. The equations to the paths are

cos-1 -- 3 cos"1 -= 0, and cos-^ + S sin" 1 -= 0,a a a a

where a is the common amplitude.

CHAPTER IX. (Pages 315320.)

4. The component velocities are ^ ae and ^ a respectively.

8. 219 days nearly.

13. The new major axis is 4a; the new eccentricity is ^

the new major axis makes an angle tan" 1 - with the original major

axis.

22.1 - e2 / V sa\

The increase in the eccentricity is yV I 1 -j

23. The direction of the blow bisects the angle between the minor

axis and the radius vector to the centre of force.

24. The axis is unaltered in direction.

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